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AI HL · AHL 2

Further Derivatives

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AHL 2.3

The chain rule

AHL 2.3

The chain rule

Target: I can pick the inner function u and use dy/dx = dy/du × du/dx to differentiate powers, roots, sin, cos, e and ln of a function. (+4 more)

By the end of this section

Key points

The chain rule

  • A composite function has one function inside another, such as \((x^2-3x)^6\) or \(\sin 4x\). There is no log law to rescue \(\sin x^3\) or \(e^{x^3}\).
  • If \(y=f(u)\) and \(u=g(x)\), then \(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\).
  • Method: choose the inner function u, write y in terms of u, find dy/du and du/dx, multiply, then replace u with its expression in x.
  • Three nested functions need three derivatives multiplied: \(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dx}\).

Results to know

  • \(\frac{d}{dx}(ax+b)^n=na(ax+b)^{n-1}\) and \(\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\).
  • \(\frac{d}{dx}\sin(ax+b)=a\cos(ax+b)\) and \(\frac{d}{dx}\cos(ax+b)=-a\sin(ax+b)\), with x in radians.
  • \(\frac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\) and \(\frac{d}{dx}\ln f(x)=\frac{f'(x)}{f(x)}\).
  • \(\frac{1}{e^{kx}}=e^{-kx}\), so its derivative is \(-ke^{-kx}\). Also \((e^x)^2=e^{2x}\), so simplify before you differentiate.

Watch for

  • The order matters: \(\sqrt{\cos x}\) has the cosine inside, \(\cos\sqrt{x}\) has the square root inside. The answers differ.
  • Do not forget du/dx. The most common mistake is \(\frac{d}{dx}\sin 4x=\cos 4x\).
  • Tangent: gradient m = dy/dx at the point, then \(y-y_1=m(x-x_1)\). Parallel lines have equal gradients. Perpendicular gradients multiply to −1.

On the calculator

Radians for trig derivatives

  • Set the calculator to radians before you evaluate sin, cos or a gradient of a trig curve.
  • A trig derivative such as \(3\cos(3x+2)\) at x = 0 is \(3\cos 2\): in degrees it gives 2.998, which is wrong.

Screenshot and video to come

Gradient at a point

  • Differentiate by hand first, then evaluate the derivative at the x value from the calculator.
  • Answers to 3 significant figures unless the question asks for an exact value such as ln 3.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Power of a quadratic

Differentiate \(y=(x^2-4x)^5\), then find the gradient at \(x=1\).

Working
  1. Choose u

    \(u=x^2-4x,\ y=u^5\)

    The inner function is the bracket.

  2. Two derivatives

    \(\frac{dy}{du}=5u^4,\ \frac{du}{dx}=2x-4\)

    Differentiate the outer power, then the inside.

  3. Multiply

    \(\frac{dy}{dx}=5u^4(2x-4)\)

    dy/dx = dy/du × du/dx.

  4. Replace u

    \(\frac{dy}{dx}=5(x^2-4x)^4(2x-4)\)

    The answer must be in x.

  5. At x = 1

    \(5(-3)^4(-2)=-810\)

    Substitute x = 1: u = −3, du/dx = −2.

Video walkthrough coming
Sine of a linear function

Differentiate \(y=\sin(3x+2)\), then find the gradient at \(x=0\) with the calculator in radians.

Working
  1. Choose u

    \(u=3x+2,\ y=\sin u\)

    The inner function is the angle.

  2. Two derivatives

    \(\frac{dy}{du}=\cos u,\ \frac{du}{dx}=3\)

    The 3 comes out in front.

  3. Replace u

    \(\frac{dy}{dx}=3\cos(3x+2)\)

    Multiply and put x back.

  4. At x = 0

    From the calculator (radians): \(3\cos 2=-1.25\)

    Check the calculator is in radian mode.

Video walkthrough coming
e to a polynomial

Differentiate \(y=e^{2x^2+x}\), then find the gradient at \(x=1\).

Working
  1. Choose u

    \(u=2x^2+x,\ y=e^u\)

    The inner function is the power.

  2. Two derivatives

    \(\frac{dy}{du}=e^u,\ \frac{du}{dx}=4x+1\)

    e^u does not change when differentiated.

  3. Replace u

    \(\frac{dy}{dx}=(4x+1)e^{2x^2+x}\)

    Write the inner derivative in front.

  4. At x = 1

    From the calculator: \(5e^3=100\)

    u = 3 and du/dx = 5 at x = 1.

Video walkthrough coming
Three nested functions

Differentiate \(y=(\ln 4x)^3\), then find the gradient at \(x=2\).

Working
  1. Three layers

    \(y=u^3,\ u=\ln v,\ v=4x\)

    Cube of a logarithm of a linear function.

  2. Three derivatives

    \(\frac{dy}{du}=3u^2,\ \frac{du}{dv}=\frac1v,\ \frac{dv}{dx}=4\)

    Multiply all three.

  3. Multiply

    \(\frac{dy}{dx}=3(\ln 4x)^2\times\frac{1}{4x}\times4\)

    Replace v with 4x and u with ln 4x.

  4. Tidy

    \(\frac{dy}{dx}=\frac{3(\ln 4x)^2}{x}\)

    The 4s cancel.

  5. At x = 2

    From the calculator: \(\frac{3(\ln 8)^2}{2}=6.49\)

    Substitute x = 2.

Video walkthrough coming
Which function is inside?

Differentiate (a) \(y=\sqrt{\cos x}\) and (b) \(y=\cos\sqrt{x}\). For (b), find the gradient at \(x=4\).

Working
  1. (a) The inner function

    \(u=\cos x,\ y=u^{1/2}\)

    The cosine is inside the square root.

  2. (a) Multiply

    \(\frac{dy}{dx}=\frac12u^{-1/2}\times(-\sin x)=\frac{-\sin x}{2\sqrt{\cos x}}\)

    dy/du × du/dx.

  3. (b) The inner function

    \(u=\sqrt{x}=x^{1/2},\ y=\cos u\)

    Now the square root is inside. The order has changed the answer.

  4. (b) Multiply

    \(\frac{dy}{dx}=-\sin u\times\frac{1}{2}x^{-1/2}=\frac{-\sin\sqrt x}{2\sqrt x}\)

    du/dx = ½x^(−1/2).

  5. (b) At x = 4

    From the calculator (radians): \(\frac{-\sin 2}{4}=-0.227\)

    √4 = 2.

Video walkthrough coming
Differentiating 1/e^(2x)

Differentiate \(y=\frac{1}{e^{2x}}\), then find the gradient at \(x=0\).

Working
  1. Rewrite

    \(y=e^{-2x}=e^u,\ u=-2x\)

    1/e^(2x) is e^(−2x). Then use the chain rule.

  2. Two derivatives

    \(\frac{dy}{du}=e^u,\ \frac{du}{dx}=-2\)

    The inner derivative is negative.

  3. Replace u

    \(\frac{dy}{dx}=-2e^{-2x}=-\frac{2}{e^{2x}}\)

    Either form is fine.

  4. At x = 0

    \(-2e^0=-2\)

    e⁰ = 1.

Video walkthrough coming
Equation of a tangent

Find the equation of the tangent to \(y=\sqrt{x^2+5}\) at the point where \(x=2\). Write it as \(y=mx+c\).

−112345123456xyy = √(x²+5)
Working
  1. Differentiate

    \(u=x^2+5,\ y=u^{1/2}:\ \frac{dy}{dx}=\frac12u^{-1/2}\times2x=\frac{x}{\sqrt{x^2+5}}\)

    Chain rule with a square root.

  2. Point and gradient

    \(x=2:\ y=3,\ m=\frac23\)

    Substitute x = 2 into the curve and the derivative.

  3. Line

    \(y-3=\frac23(x-2)\)

    y − y₁ = m(x − x₁).

  4. Rearrange

    \(y=\frac23x+\frac53\)

    So c = 5/3 = 1.67.

Video walkthrough coming
Parallel and perpendicular tangents

The curve is \(y=\ln(4x-3)\). (a) Find the point where the tangent is parallel to \(y=2x\). (b) Find the x-coordinate of the point where the tangent is perpendicular to \(y=-2x+1\).

Working
  1. Gradient function

    \(u=4x-3:\ \frac{dy}{dx}=\frac1u\times4=\frac{4}{4x-3}\)

    Chain rule for ln of a linear function.

  2. (a) Parallel

    \(\frac{4}{4x-3}=2\Rightarrow 4x-3=2\Rightarrow x=1.25\)

    Parallel lines have the same gradient: 2.

  3. (a) The point

    \(y=\ln(2)=0.693\), point \((1.25,\ 0.693)\)

    Substitute x into the curve.

  4. (b) Perpendicular

    \(\frac{4}{4x-3}=\frac12\Rightarrow 4x-3=8\Rightarrow x=2.75\)

    Perpendicular to gradient −2 means gradient ½ (the negative reciprocal).

Video walkthrough coming