Mini-Lesson

Differentiation

4 missions. Guess, learn, then check. Your place is saved.

Mission 1Speed needs two moments. So what is your speed at one instant?
Mission 2Fold a box from a 30 cm square card: what corner cut holds the most?
Mission 3How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?
Mission 4A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Speed needs two moments. So what is your speed at one instant?

Make a guess: Speed needs two moments. So what is your speed at one instant?

Differentiation · first principles

Chords close in on the tangent

The gradient of a curve at a point is the gradient of the tangent there.
A chord joins $P (x, f (x))$ P(x, f(x)) to a nearby point $Q (x + h, f (x + h))$ Q(x+h, f(x+h)). Its gradient is

\frac{f ( x + h ) - f ( x )}{h}

As $h \to 0$ h → 0, $Q$ Q slides towards $P$ P and the chord turns into the tangent. The tangent's gradient is the limit of the chord gradients.
We write it $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x): the rate of change of $y$ y with respect to $x$ x.

Worked example Differentiation · first principles

Chord gradients on $y = x^{2}$ y = x²

Question

P (3, 9)

lies on

y = x^{2}

. Find the gradients of the chords from

P

to the points with

x = 3 + h

for

h = 1

,

0.1

and

0.01

, and hence estimate the gradient at

P

.

1
Use the chord gradient $\frac{( 3 + h ) ^{2} - 9}{h}$ ((3+h)² − 9)/h for each $h$ h.
2
$h = 1$ h = 1: $\frac{16 - 9}{1} = 7$ (16 − 9)/1 = 7
3
$h = 0.1$ h = 0.1: $\frac{9.61 - 9}{0.1} = 6.1$ (9.61 − 9)/0.1 = 6.1
4
$h = 0.01$ h = 0.01: $\frac{9.0601 - 9}{0.01} = 6.01$ (9.0601 − 9)/0.01 = 6.01
5
The gradients approach 6. In fact $\frac{( 3 + h ) ^{2} - 9}{h} = 6 + h$ ((3+h)² − 9)/h = 6 + h, which tends to 6 as $h \to 0$ h → 0.

Answer

the chord gradients are 7, 6.1 and 6.01; the gradient at

P

is 6.

⚠️ Watch out

putting

h = 0

straight into

\frac{( 3 + h ) ^{2} - 9}{h}

gives

\frac{0}{0}

. Simplify first, then let

h \to 0

.

Quick check

?On

y = x^{3}

, what is the gradient of the chord from

x = 2

to

x = 2.1

?

Differentiation · first principles

The derivative as a limit

Definition:

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Method: expand $f (x + h)$ f(x+h), subtract $f (x)$ f(x), divide by $h$ h, then let $h \to 0$ h → 0.
For $f (x) = x^{2}$ f(x) = x²: $\frac{( x + h ) ^{2} - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h} = 2 x + h$ ((x+h)² − x²)/h = (2xh + h²)/h = 2x + h, which tends to $2 x$ 2x. So $\frac{d}{d x} (x^{2}) = 2 x$ d/dx(x²) = 2x.
You may see $δ x$ δ x in place of $h$ h: $\frac{d y}{d x} = lim_{δ x \to 0} \frac{δy}{δ x}$ dy/dx = lim_(δ x → 0) (δ y)/(δ x).

Calculate

?For

f (x) = 2 x^{2} - 3 x

, show that

\frac{f ( x + h ) - f ( x )}{h} = 4 x + 2 h - 3

. Hence find the gradient of the curve

y = f (x)

at

x = 5

. Give an exact answer.

Calculate

?A ball rolls down a ramp so that it has travelled

s = 0.6 t^{2}

metres after

t

seconds. Show that its average speed from

t = 3

to

t = 3 + h

is

3.6 + 0.6 h

m/s. Hence find its speed at the instant

t = 3

. Give an exact answer.

m/s

Mission check

?Speed needs two moments. So what is your speed at one instant?

Mission 2

Fold a box from a 30 cm square card: what corner cut holds the most?

Make a guess: Fold a box from a 30 cm square card: what corner cut holds the most?

Differentiation · stationary points

The power rule

For any rational $n$ n:

\frac{d}{d x} (a x^{n}) = an x^{n - 1}

Differentiate a sum or difference term by term. A constant term differentiates to 0.
First rewrite as powers of $x$ x: $x = x^{\frac{1}{2}}$ √x = x^(1/2), $\frac{1}{x ^{2}} = x^{- 2}$ 1/x² = x⁻², $\frac{3}{x} = 3 x^{- \frac{1}{2}}$ 3/(√x) = 3x^(−1/2). Expand brackets and split fractions over a single power.
Notation: $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x). Differentiating again gives the second derivative $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx² or $f^{''} (x)$ f''(x).

Worked example Differentiation · stationary points

Rewriting before differentiating

Question

Given

y = \frac{x ^{2} + 6}{x}

,

x > 0

, find

\frac{d y}{d x}

and the gradient of the curve at

x = 4

.

1
Split the fraction and write as powers: $y = x^{\frac{3}{2}} + 6 x^{- \frac{1}{2}}$ y = x^(3/2) + 6x^(−1/2)
2
Multiply by the power, then reduce the power by 1: $\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} - 3 x^{- \frac{3}{2}}$ dy/dx = 3/2x^(1/2) − 3x^(−3/2)
3
Substitute $x = 4$ x = 4: $\frac{3}{2} (2) - 3 \times \frac{1}{8} = 3 - \frac{3}{8}$ 3/2(2) − 3 × 1/8 = 3 − 3/8

Answer

\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} - 3 x^{- \frac{3}{2}}

; the gradient at

x = 4

is

\frac{21}{8} = 2.625

.

⚠️ Watch out

you cannot differentiate the top and bottom separately. Rewrite as a sum of powers first.

Quick check

?Find

\frac{d y}{d x}

when

y = \frac{8}{x ^{3}}

.

Differentiation · stationary points

Tangents and normals

The gradient of the curve at $x = a$ x = a is $m = f^{'} (a)$ m = f'(a). The tangent there has gradient $m$ m.
The normal is perpendicular to the tangent, so its gradient is $- \frac{1}{m}$ −1/m.
Use $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) with the point $(a, f (a))$ (a, f(a)) on the curve.
Example: $y = x^{2}$ y = x² at $(3, 9)$ (3, 9) has gradient 6, so the tangent is $y - 9 = 6 (x - 3)$ y − 9 = 6(x − 3), that is $y = 6 x - 9$ y = 6x − 9.

Calculate

?The curve

C

has equation

y = x^{3} - 3 x^{2} + 2

. The normal to

C

at the point where

x = 3

meets the

x

-axis at

Q

. Find the

x

-coordinate of

Q

. Give an exact answer.

Differentiation · stationary points

Stationary points and their nature

$f$ f is increasing where $f^{'} (x) > 0$ f'(x) > 0 and decreasing where $f^{'} (x) < 0$ f'(x) < 0.
At a stationary point $\frac{d y}{d x} = 0$ dy/dx = 0. Solve this to find the $x$ x-coordinates.
Classify with the second derivative: $\frac{d ^{2} y}{d x ^{2}} < 0$ d²y/dx² < 0 gives a maximum, $\frac{d ^{2} y}{d x ^{2}} > 0$ d²y/dx² > 0 a minimum. If it is 0, check the sign of $\frac{d y}{d x}$ dy/dx either side.
Optimisation: write the quantity in terms of one variable, differentiate, set the derivative to 0, classify, then answer in context.

Worked example Differentiation · stationary points

Finding and classifying stationary points

Question

Find the stationary points of

y = x^{3} - 6 x^{2} + 9 x + 1

and determine their nature.

1
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3)
2
Set to zero: $x = 1$ x = 1 or $x = 3$ x = 3, giving $y = 5$ y = 5 and $y = 1$ y = 1
3
Differentiate again: $\frac{d ^{2} y}{d x ^{2}} = 6 x - 12$ d²y/dx² = 6x − 12
4
At $x = 1$ x = 1: $6 - 12 = - 6 < 0$ 6 − 12 = −6 < 0, a maximum. At $x = 3$ x = 3: $18 - 12 = 6 > 0$ 18 − 12 = 6 > 0, a minimum.

Answer

(1, 5)

is a maximum and

(3, 1)

is a minimum.

⚠️ Watch out

give both coordinates. Many students find

x = 1

and

x = 3

and stop, losing the

y

-values.

Sort it

?For

f (x) = x^{3} - 12 x

, decide whether

f

is increasing or decreasing at each value of

x

.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?A closed cylindrical can has volume 500 cm³ and radius

r

cm. Show that its total surface area is

A = 2 π r^{2} + \frac{1000}{r}

. Use calculus to find the value of

r

that makes

A

a minimum. Give your answer to 3 significant figures.

cm

Mission check

?Fold a box from a 30 cm square card: what corner cut holds the most?

Mission 3

How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?

Make a guess: How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?

Differentiation · chain, product and quotient rules

Standard derivatives (A2 only)

$y$ y	$\frac{d y}{d x}$ dy/dx
$e^{k x}$ e^kx	$k e^{k x}$ ke^kx
$ln x$ ln x (and $ln k x$ ln kx)	$\frac{1}{x}$ 1/x
$sin k x$ sin kx	$k cos k x$ kcos kx
$cos k x$ cos kx	$- k sin k x$ −ksin kx
$tan k x$ tan kx	$k sec^{2} k x$ ksec² kx

Angles must be in radians. The trig results are false in degrees.
$ln k x = ln k + ln x$ ln kx = ln k + ln x, so the constant $k$ k disappears when you differentiate.

Differentiation · chain, product and quotient rules

The chain rule (A2 only)

For a function of a function, $y = f (u)$ y = f(u) where $u = g (x)$ u = g(x):

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

In words: differentiate the outer function, leave the inside alone, then multiply by the derivative of the inside.
Rates multiply along the chain: if $u$ u changes 2 times as fast as $x$ x, and $y$ y 3 times as fast as $u$ u, then $y$ y changes 6 times as fast as $x$ x.

Worked example Differentiation · chain, product and quotient rules

Using the chain rule (A2 only)

Question

Differentiate

y = (2 x^{3} - 1)^{5}

.

1
Name the inside: $u = 2 x^{3} - 1$ u = 2x³ − 1, so $y = u^{5}$ y = u⁵
2
Differentiate each link: $\frac{d y}{d u} = 5 u^{4}$ dy/du = 5u⁴ and $\frac{d u}{d x} = 6 x^{2}$ du/dx = 6x²
3
Multiply and put $u$ u back: $\frac{d y}{d x} = 5 (2 x^{3} - 1)^{4} \times 6 x^{2}$ dy/dx = 5(2x³ − 1)⁴ × 6x²

Answer

\frac{d y}{d x} = 30 x^{2} (2 x^{3} - 1)^{4}

⚠️ Watch out

5 (2 x^{3} - 1)^{4}

alone forgets to multiply by the derivative of the inside. Expanding the bracket first works but wastes time.

Quick check

?Find

\frac{d y}{d x}

when

y = e^{3 x^{2}}

.

Differentiation · chain, product and quotient rules

Product and quotient rules (A2 only)

Product rule: if $y = uv$ y = uv, then $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ dy/dx = udv/dx + vdu/dx.
Quotient rule: if $y = \frac{u}{v}$ y = u/v, then $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ dy/dx = (vdu/dx − udv/dx)/v². The order on top matters.
The quotient rule on $\frac{s i n x}{c o s x}$ (sin x)/(cos x) gives $\frac{c o s ^{2} x + s i n ^{2} x}{c o s ^{2} x} = sec^{2} x$ (cos² x + sin² x)/(cos² x) = sec² x.
The chain or quotient rule gives $\frac{d}{d x} sec x = sec x tan x$ d/dxsec x = sec xtan x, $\frac{d}{d x} cosec x = - cosec x cot x$ d/dxcosec x = −cosec xcot x, $\frac{d}{d x} cot x = - cosec^{2} x$ d/dxcot x = −cosec² x.

Worked example Differentiation · chain, product and quotient rules

Product rule and a stationary point (A2 only)

Question

Find the

x

-coordinates of the stationary points of

y = x^{2} e^{3 x}

.

1
Set $u = x^{2}$ u = x² and $v = e^{3 x}$ v = e^3x: $\frac{d u}{d x} = 2 x$ du/dx = 2x and $\frac{d v}{d x} = 3 e^{3 x}$ dv/dx = 3e^3x
2
Product rule: $\frac{d y}{d x} = 3 x^{2} e^{3 x} + 2 x e^{3 x}$ dy/dx = 3x²e^3x + 2xe^3x
3
Factorise: $\frac{d y}{d x} = x e^{3 x} (3 x + 2)$ dy/dx = xe^3x(3x + 2)
4
Set to zero: $e^{3 x} > 0$ e^3x > 0 for all $x$ x, so $x = 0$ x = 0 or $3 x + 2 = 0$ 3x + 2 = 0

Answer

x = 0

and

x = - \frac{2}{3}

⚠️ Watch out

never divide by

x

to "cancel" it: that loses the solution

x = 0

. Factorise instead.

Calculate

?Find the gradient of the curve

y = \frac{2 x - 1}{x + 3}

at the point where

x = 1

. Give your answer as a decimal.

Quick check

?Find

\frac{d y}{d x}

when

y = ln 3 x + sin 4 x

(

x > 0

, in radians).

Calculate

?The curve

y = (x - 2) e^{2 x}

has one stationary point. Find its

x

-coordinate. Give your answer as a decimal.

Mission check

?How do you differentiate

e^{s i n x}

, a function inside a function?

Mission 4

A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Make a guess: A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Differentiation · rates, implicit and parametric

Connected rates of change (A2 only)

When two quantities are linked by a formula, their rates are linked by the chain rule:

\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}

Ripple: $A = π r^{2}$ A = π r² and $\frac{d r}{d t} = 2$ dr/dt = 2 cm/s. At $r = 5$ r = 5: $\frac{d A}{d t} = 2 π r \times 2 = 20 π \approx 62.8$ dA/dt = 2π r × 2 = 20π ≈ 62.8 cm²/s.
CCEA (A2 1) does not name connected rates of change: here it is an application of the chain rule you must know.
To construct a differential equation, turn words into a rate: " $N$ N grows at a rate proportional to $N$ N" becomes $\frac{d N}{d t} = k N$ dN/dt = kN, with $k > 0$ k > 0. "Decreases" needs a minus sign.

Calculate

?Water is poured into an inverted cone at a constant 6 cm³ per second. When the water is

h

cm deep, the surface radius is

\frac{h}{2}

cm. Show that

V = \frac{π h ^{3}}{12}

. Find the rate at which the depth is rising when

h = 4

. Give your answer to 3 significant figures.

cm/s

Quick check

?In a simple model, the rate of decrease of the radius

r

of a mint is inversely proportional to the square of the radius. Which differential equation fits (

k

a positive constant,

t

time)?

Differentiation · rates, implicit and parametric

Implicit differentiation (A2 only)

In an equation like $x^{2} + x y + y^{2} = 7$ x² + xy + y² = 7, $y$ y is not the subject. Differentiate every term with respect to $x$ x.
A term in $y$ y needs the chain rule: $\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}$ d/dx(y³) = 3y²dy/dx.
A product needs the product rule: $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$ d/dx(xy) = xdy/dx + y.
Then collect the $\frac{d y}{d x}$ dy/dx terms on one side and factorise.
CCEA also asks for $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx²: differentiate your $\frac{d y}{d x}$ dy/dx again, implicitly, and substitute.

Worked example Differentiation · rates, implicit and parametric

Gradient of an implicit curve (A2 only)

Question

Find the gradient of the curve

x^{2} + x y + y^{2} = 7

at the point

(1, 2)

.

1
Check the point: $1 + 2 + 4 = 7$ 1 + 2 + 4 = 7, so $(1, 2)$ (1, 2) is on the curve.
2
Differentiate term by term: $2 x + (x \frac{d y}{d x} + y) + 2 y \frac{d y}{d x} = 0$ 2x + (xdy/dx + y) + 2ydy/dx = 0
3
Collect: $\frac{d y}{d x} (x + 2 y) = - (2 x + y)$ dy/dx(x + 2y) = −(2x + y), so $\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y}$ dy/dx = −(2x + y)/(x + 2y)
4
Substitute $(1, 2)$ (1, 2): $\frac{d y}{d x} = - \frac{4}{5}$ dy/dx = −4/5

Answer

the gradient at

(1, 2)

is

- \frac{4}{5} = - 0.8

.

⚠️ Watch out

differentiating

x y

as just

\frac{d y}{d x}

(no product rule) gives

- \frac{2}{5}

, a common lost mark.

Calculate

?For the circle

x^{2} + y^{2} = 25

, find the value of

\frac{d ^{2} y}{d x ^{2}}

at the point

(3, 4)

. Give your answer as a decimal.

Differentiation · rates, implicit and parametric

Parametric differentiation (A2 only)

If $x$ x and $y$ y are both given in terms of a parameter $t$ t, use the chain rule:

\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}

The answer is in terms of $t$ t. For a tangent or normal at a given $t$ t, find the point $(x, y)$ (x, y) and the gradient from the same $t$ t.
Second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \div \frac{d x}{d t}$ d²y/dx² = d/dt(dy/dx) ÷ dx/dt.

Worked example Differentiation · rates, implicit and parametric

Tangent to a parametric curve (A2 only)

Question

A curve has parametric equations

x = t^{2} + 1

,

y = t^{3} - 3 t

. Find the equation of the tangent at the point where

t = 2

.

1
Find the point: $x = 5$ x = 5, $y = 8 - 6 = 2$ y = 8 − 6 = 2
2
Differentiate each: $\frac{d x}{d t} = 2 t = 4$ dx/dt = 2t = 4 and $\frac{d y}{d t} = 3 t^{2} - 3 = 9$ dy/dt = 3t² − 3 = 9 at $t = 2$ t = 2
3
Divide: $\frac{d y}{d x} = \frac{9}{4}$ dy/dx = 9/4
4
Tangent: $y - 2 = \frac{9}{4} (x - 5)$ y − 2 = 9/4(x − 5), so $4 y = 9 x - 37$ 4y = 9x − 37

Answer

y = \frac{9}{4} x - \frac{37}{4}

, or

9 x - 4 y - 37 = 0

⚠️ Watch out

dividing the other way,

\frac{d x}{d t} \div \frac{d y}{d t}

, gives

\frac{4}{9}

: that is

\frac{d x}{d y}

.

Calculate

?For the curve

x = t^{2} + 1

,

y = t^{3} - 3 t

from the Example above, find the value of

\frac{d ^{2} y}{d x ^{2}}

at the point where

t = 2

. Give your answer as a decimal.

Mission check

?A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Recap

Big ideas

The derivative $\frac{d y}{d x} = f^{'} (x)$ dy/dx = f'(x) is the gradient of the tangent and the rate of change of $y$ y with respect to $x$ x.
It is the limit of chord gradients: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ f'(x) = lim_(h → 0) (f(x+h) − f(x))/h. Expand, subtract, divide by $h$ h, then let $h \to 0$ h → 0.
$\frac{d}{d x} (a x^{n}) = an x^{n - 1}$ d/dx(axⁿ) = anxⁿ⁻¹ for rational $n$ n: rewrite roots and fractions as powers first.
Tangent gradient $m = f^{'} (a)$ m = f'(a); normal gradient $- \frac{1}{m}$ −1/m; then $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁).
Stationary points solve $\frac{d y}{d x} = 0$ dy/dx = 0; $\frac{d ^{2} y}{d x ^{2}} < 0$ d²y/dx² < 0 means a maximum and $> 0$ > 0 a minimum.
Learn the standard derivatives of $e^{k x}$ e^kx, $ln x$ ln x, $sin k x$ sin kx, $cos k x$ cos kx, $tan k x$ tan kx, with $x$ x in radians.
Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ dy/dx = dy/du × du/dx. Outer, keep the inside, times the inner derivative.
Product: $u \frac{d v}{d x} + v \frac{d u}{d x}$ udv/dx + vdu/dx. Quotient: $\frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ (vdu/dx − udv/dx)/v².
Connected rates: $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ dV/dt = dV/dr × dr/dt. Words like "proportional to" and "decreases" become a differential equation.
Implicit: differentiate every term with respect to $x$ x; $y^{n}$ yⁿ gives $n y^{n - 1} \frac{d y}{d x}$ nyⁿ⁻¹dy/dx and $x y$ xy needs the product rule.
Parametric: $\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}$ dy/dx = dy/dt ÷ dx/dt.
For $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx², differentiate $\frac{d y}{d x}$ dy/dx with respect to $t$ t and divide by $\frac{d x}{d t}$ dx/dt again.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Differentiation for CCEA GCE Mathematics (2210). 🎉

Mission 1Speed needs two moments. So what is your speed at one instant?
Mission 2Fold a box from a 30 cm square card: what corner cut holds the most?
Mission 3How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?
Mission 4A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

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→ Back to A-Level Mathematics · CCEA

Specification points and sources for this lesson are listed at the bottom of the page.

Differentiation

Speed needs two moments. So what is your speed at one instant?

Chords close in on the tangent

Chord gradients on y=x2y = x²

The derivative as a limit

Fold a box from a 30 cm square card: what corner cut holds the most?

The power rule

Rewriting before differentiating

Tangents and normals

Stationary points and their nature

Finding and classifying stationary points

How do you differentiate esinxe^(sin x), a function inside a function?

Standard derivatives (A2 only)

The chain rule (A2 only)

Using the chain rule (A2 only)

Product and quotient rules (A2 only)

Product rule and a stationary point (A2 only)

A balloon's radius grows 1 cm/s. Why does its volume grow ever faster?

Connected rates of change (A2 only)

Implicit differentiation (A2 only)

Gradient of an implicit curve (A2 only)

Parametric differentiation (A2 only)

Tangent to a parametric curve (A2 only)

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Chord gradients on $y = x^{2}$ y = x²

How do you differentiate $e^{s i n x}$ e^(sin x), a function inside a function?