How do you find the gradient of a tangent to an exponential curve?

Differentiate the function and substitute the x value of the point; the derivative there is the gradient of the tangent. For y equal to e to the 2x, the derivative is 2 e to the 2x, so at x equal to 0 the gradient is 2 and the tangent line is y equals 2x plus 1.

Why is e to the x special in calculus?

The number e is the unique base for which the exponential function equals its own derivative, so d/dx of e to the x equals e to the x. This makes e the natural base for describing continuous growth and decay, where the rate of change is proportional to the amount present.

How do you find the gradient of a tangent to an exponential curve?

Differentiate and substitute the point. For y equal to e to the 2x the derivative is 2 e to the 2x, so at x equal to 0 the gradient is 2 and the tangent is y equals 2x plus 1.

Why is e to the x special in calculus?

e is the unique base for which the function equals its own derivative, so d/dx of e to the x equals e to the x. This makes e the natural base for continuous growth and decay.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Differentiation of e^x

Q: What is the derivative of e to the x?

The exponential function is its own derivative: d/dx of e to the x equals e to the x. The gradient at any point equals the height of the curve there; at (0,1) the gradient is 1.

Q: How do you differentiate e to the power f of x?

Use the chain rule: d/dx of e to the f(x) equals f prime of x times e to the f(x). Differentiate the exponent to get f prime of x, then multiply the exponential by that factor, e.g. d/dx of e to the x squared equals 2x times e to the x squared.

Q: How do you differentiate e to the kx?

For a linear exponent, d/dx of e to the kx equals k times e to the kx. So d/dx of e to the 3x equals 3 e to the 3x. A constant multiple in front is unchanged.

Q: How do you differentiate x times e to the x?

Product rule: u prime v plus u v prime with u equal to x and v equal to e to the x gives e to the x plus x e to the x, which is (x plus 1) e to the x.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), differentiating \(e^x\) uses two rules: \(\dfrac{d}{dx}e^{x}=e^{x}\) and, for a composite, \(\dfrac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\). A constant multiple is carried through, a product or quotient with \(e^{x}\) uses the product or quotient rule, and the derivative at a point gives the gradient of the tangent there.

The exponential function \(y=e^{x}\) is the one function that is its own derivative: \(\dfrac{d}{dx}e^{x}=e^{x}\). Geometrically, the gradient of the curve at any point equals its height there, so at \((0,1)\) the gradient is \(1\). The number \(e\approx 2.718\) is the unique base with this property.

For a composite \(e^{f(x)}\), the chain rule multiplies by the derivative of the exponent: \(\dfrac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\). When the exponent is linear, \(f(x)=kx\), this gives \(\dfrac{d}{dx}e^{kx}=k\,e^{kx}\) — for example \(\dfrac{d}{dx}e^{3x}=3e^{3x}\).

An \(e^{x}\) term multiplied by, or divided by, another function is differentiated with the product rule \((uv)'=u'v+uv'\) or the quotient rule \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}\). Evaluating the derivative at \(x=a\) gives the gradient of the tangent at that point, from which the tangent line \(y-y_1=m(x-x_1)\) follows.

Key idea. \(\dfrac{d}{dx}e^{x}=e^{x}\); by the chain rule \(\dfrac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\). Never drop the factor \(f'(x)\), and remember \(e^{x}\) is not a power of \(x\).

\(y=e^{x}\): the tangent at \((0,1)\) is \(y=x+1\), so the gradient is \(1\)

\(y=e^{2x}\) is twice as steep at \((0,1)\); the tangent is \(y=2x+1\), gradient \(2\)

The base rule — the exponential is its own derivative:

\[\frac{d}{dx}e^{x}=e^{x}\]

\frac{d}{d x} e^{x} = e^{x}

The chain rule for a composite exponential (multiply by \(f'(x)\)):

\[\frac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)} \qquad \frac{d}{dx}e^{kx}=k\,e^{kx}\]

\frac{d}{d x} e^{f (x)} = f^{'} (x) e^{f (x)}

The product and quotient rules, used when \(e^{x}\) multiplies or divides another function:

\[(uv)'=u'v+uv' \qquad \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}\]

(u v)^{'} = u^{'} v + u v^{'}

Gradient of a tangent. The derivative at \(x=a\) is the gradient there; e.g. for \(y=e^{2x}\), \(\dfrac{dy}{dx}=2e^{2x}\), so at \(x=0\) the gradient is \(2\) and the tangent is \(y=2x+1\).

How to differentiate an expression involving \(e^{x}\)

Spot the form. A bare \(e^{x}\) (or \(a\,e^{x}\)) uses \(\dfrac{d}{dx}e^{x}=e^{x}\); an exponent that is a function of \(x\) needs the chain rule.
Chain rule for \(e^{f(x)}\). Differentiate the exponent to get \(f'(x)\), then write \(f'(x)\,e^{f(x)}\). For \(e^{kx}\) this is just \(k\,e^{kx}\).
Product or quotient. If \(e^{x}\) multiplies or divides another function, apply \((uv)'=u'v+uv'\) or \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}\), then simplify and factorise.
Gradient or tangent. To find a gradient, substitute the \(x\)-value into the derivative; for a tangent line use \(y-y_1=m(x-x_1)\) with that gradient and the point.

Simplify-first shortcut. A quotient such as \(\dfrac{e^{2x}+1}{e^{2x}}\) simplifies to \(1+e^{-2x}\) before differentiating — often quicker than the quotient rule.

Example 1 — constant multiple and \(e^{kx}\)

Differentiate \(y=4e^{x}\) and \(y=e^{5x}\).

Solution

Carry the constant; for \(e^{5x}\) multiply by the inner derivative \(5\).

\(\dfrac{d}{dx}(4e^{x})\)	\(=\)	\(4e^{x}\)
\(\dfrac{d}{dx}(e^{5x})\)	\(=\)	\(5e^{5x}\)

5 e^{5 x}

Example 2 — chain rule \(e^{f(x)}\)

Differentiate \(y=e^{x^{2}-x}\).

Solution

Inner \(f=x^{2}-x\), so \(f'=2x-1\).

\(\dfrac{dy}{dx}\)	\(=\)	\(f'(x)\,e^{f(x)}\)
	\(=\)	\((2x-1)e^{x^{2}-x}\)

(2 x - 1) e^{x^{2} - x}

Example 3 — product rule

Differentiate \(y=x\,e^{x}\).

Solution

Product rule with \(u=x\), \(v=e^{x}\).

\(y'\)	\(=\)	\((1)e^{x}+x(e^{x})\)
	\(=\)	\((x+1)e^{x}\)

(x + 1) e^{x}

Example 4 — gradient of a tangent

Find the gradient of \(y=e^{2x}\) at \((0,1)\), and the tangent line.

Solution

Differentiate, then substitute \(x=0\).

\(\dfrac{dy}{dx}\)	\(=\)	\(2e^{2x}\)
\(\text{at } x=0\)	\(=\)	\(2e^{0}=2\)

Tangent: \(y-1=2(x-0)\), so \(y=2x+1\).

y = 2 x + 1

Common pitfalls

Do not drop the factor \(f'(x)\). By the chain rule \(\dfrac{d}{dx}e^{5x}=5e^{5x}\), not \(e^{5x}\). The inner derivative always multiplies the exponential.

\(e^{x}\) is not a power of \(x\). The power rule does not apply: \(\dfrac{d}{dx}e^{x}=e^{x}\), never \(x\,e^{x-1}\) and never \(xe^{x}\).

Multiply by the derivative of the exponent, not the exponent itself. \(\dfrac{d}{dx}e^{x^{2}}=2x\,e^{x^{2}}\), not \(x^{2}e^{x^{2}}\).

Frequently asked questions

What is the derivative of e to the x?

It is itself: \(\dfrac{d}{dx}e^{x}=e^{x}\). The gradient at any point equals the height of the curve there, so at \((0,1)\) the gradient is \(1\).

How do you differentiate e to the power f of x?

Chain rule: \(\dfrac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\). Differentiate the exponent, then multiply the exponential by that factor.

How do you differentiate e to the kx?

\(\dfrac{d}{dx}e^{kx}=k\,e^{kx}\). For example \(\dfrac{d}{dx}e^{3x}=3e^{3x}\); a constant in front is carried through.

How do you differentiate x times e to the x?

Product rule: \((1)e^{x}+x(e^{x})=(x+1)e^{x}\).

How do you find the gradient of a tangent?

Differentiate and substitute the point. For \(y=e^{2x}\), \(\dfrac{dy}{dx}=2e^{2x}\), so at \(x=0\) the gradient is \(2\) and the tangent is \(y=2x+1\).

Why is e special in calculus?

\(e\) is the unique base for which the function equals its own derivative, making it the natural base for continuous growth and decay.

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