Chain rule practice set
Twelve calculator-free questions. Questions 1 to 11 ask you to differentiate; question 12 puts it to work on a tangent.
The rule. If \(y = f(g(x))\), then
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x). \]Name the inside function \(g(x)\) first. Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Questions
Find \(\dfrac{dy}{dx}\) for each of the following.
- \(y = (3x + 2)^5\)
- \(y = (x^2 - 4)^3\)
- \(y = \sqrt{2x + 1}\)
- \(y = \dfrac{1}{(4x - 3)^2}\)
- \(y = e^{3x^2}\)
- \(y = \ln(5x - 1)\)
- \(y = \sin(4x)\)
- \(y = \cos(x^3)\)
- \(y = \sin^2 x\)
- \(y = \ln(x^2 + 1)\)
- \(y = e^{\sin x}\)
- Find the equation of the tangent to \(y = (2x - 3)^4\) at the point where \(x = 2\).
Worked solutions
1.
Inside: \(u = 3x + 2\), so \(u' = 3\).
\[ \frac{dy}{dx} = 5(3x + 2)^4 \cdot 3 \]\(\dfrac{dy}{dx} = 15(3x + 2)^4\)
2.
Inside: \(u = x^2 - 4\), so \(u' = 2x\).
\[ \frac{dy}{dx} = 3(x^2 - 4)^2 \cdot 2x \]\(\dfrac{dy}{dx} = 6x(x^2 - 4)^2\)
3.
Write it as a power: \(y = (2x + 1)^{1/2}\). Inside: \(u = 2x + 1\), so \(u' = 2\).
\[ \frac{dy}{dx} = \tfrac{1}{2}(2x + 1)^{-1/2} \cdot 2 = (2x + 1)^{-1/2} \]\(\dfrac{dy}{dx} = \dfrac{1}{\sqrt{2x + 1}}\)
4.
Write it as a power: \(y = (4x - 3)^{-2}\). Inside: \(u = 4x - 3\), so \(u' = 4\).
\[ \frac{dy}{dx} = -2(4x - 3)^{-3} \cdot 4 \]\(\dfrac{dy}{dx} = -\dfrac{8}{(4x - 3)^3}\)
5.
Inside: \(u = 3x^2\), so \(u' = 6x\). The derivative of \(e^u\) is \(e^u\).
\(\dfrac{dy}{dx} = 6x\,e^{3x^2}\)
6.
Inside: \(u = 5x - 1\), so \(u' = 5\). The derivative of \(\ln u\) is \(\dfrac{1}{u}\).
\(\dfrac{dy}{dx} = \dfrac{5}{5x - 1}\)
7.
Inside: \(u = 4x\), so \(u' = 4\). The derivative of \(\sin u\) is \(\cos u\).
\(\dfrac{dy}{dx} = 4\cos(4x)\)
8.
Inside: \(u = x^3\), so \(u' = 3x^2\). The derivative of \(\cos u\) is \(-\sin u\).
\(\dfrac{dy}{dx} = -3x^2\sin(x^3)\)
9.
\(\sin^2 x\) means \((\sin x)^2\). Inside: \(u = \sin x\), so \(u' = \cos x\).
\[ \frac{dy}{dx} = 2\sin x \cdot \cos x \]\(\dfrac{dy}{dx} = 2\sin x\cos x\)
This is also \(\sin(2x)\), if you know the double angle identity.
10.
Inside: \(u = x^2 + 1\), so \(u' = 2x\).
\(\dfrac{dy}{dx} = \dfrac{2x}{x^2 + 1}\)
11.
Inside: \(u = \sin x\), so \(u' = \cos x\).
\(\dfrac{dy}{dx} = \cos x \, e^{\sin x}\)
12.
Point: when \(x = 2\), \(y = (2 \cdot 2 - 3)^4 = 1^4 = 1\), so the point is \((2, 1)\).
Gradient: inside \(u = 2x - 3\), \(u' = 2\).
\[ \frac{dy}{dx} = 4(2x - 3)^3 \cdot 2 = 8(2x - 3)^3 \]At \(x = 2\): \(\dfrac{dy}{dx} = 8(1)^3 = 8\).
Tangent: \(y - 1 = 8(x - 2)\).
\(y = 8x - 15\)