7.7 The product, quotient and chain rules — Practise with help

1. $\dfrac{dy}{dx} = 2\cos 2x$.
2. $\dfrac{dy}{dx} = 5e^{5x}$.
3. $\dfrac{dy}{dx} = 4x^3$.

If question 2 was hard, revise the standard derivatives in Section 7.3.

Take $u = x^3$ and $v = \cos x$, so $\dfrac{du}{dx} = 3x^2$ and $\dfrac{dv}{dx} = -\sin x$.

Write $u\dfrac{dv}{dx} + v\dfrac{du}{dx} = x^3(-\sin x) + (\cos x)(3x^2)$.

The first term keeps the power and the second keeps the cosine. Both terms appear, because the product rule has two parts.

Take $u = x^2$ and $v = e^{2x}$. The derivative of $v$ is $2e^{2x}$, from the standard result for $e^{kx}$.

Write $x^2(2e^{2x}) + e^{2x}(2x)$, then take out the common factor $2xe^{2x}$.

Taking out the common factor makes the answer tidy, and shows the derivative is zero only at $x = 0$ and $x = -1$ where the exponential is never zero.

Take $u = x$ and $v = x + 1$, so $\dfrac{du}{dx} = 1$ and $\dfrac{dv}{dx} = 1$.

Write $\dfrac{(x + 1)(1) - x(1)}{(x + 1)^2}$, then simplify the numerator.

The numerator simplifies to $1$, so the derivative is always positive and the curve rises on each side of the break at $x = -1$.

Take $u = e^x$ and $v = x$. Both derivatives are simple, but the order in the numerator matters.

Write $\dfrac{x \times e^x - e^x \times 1}{x^2}$, then take out the common factor $e^x$.

The derivative is zero where $x = 1$, so the curve has a stationary point there.

Let $u = 2x - 3$, so $y = u^4$. Then multiply the derivative of the outside by the derivative of the inside.

$\dfrac{dy}{du} = 4u^3$ and $\dfrac{du}{dx} = 2$, so the product is $4(2x - 3)^3 \times 2$.

The inside derivative $2$ appears in front. Leaving it out would give an answer that is too small by a factor of $2$.

The outside function is cosine and the inside function is $3x^2$. Remember the minus sign in the derivative of cosine.

$\dfrac{dy}{dx} = -\sin(3x^2) \times \dfrac{d}{dx}(3x^2)$. Differentiate the inside next.

Both the minus sign from the cosine and the inside derivative $6x$ appear in the answer.