7.3 Differentiating powers, exponentials, trigonometric functions and logarithms — Practise with help

1. $\sqrt{x} = x^{1/2}$.
2. $\dfrac{1}{x^4} = x^{-4}$.
3. $\left(x^{1/2}\right)^2 = x^{1} = x$.

If any of these was hard, revise the laws of indices in Chapter 2 before you continue.

Multiply by the old power, which is $6 \times 5$, then reduce the power by one.

Write $\dfrac{dy}{dx} = 6 \times 5x^4$.

The power has fallen from $5$ to $4$, and the coefficient has grown from $6$ to $30$.

The power is the fraction $n = \dfrac{1}{3}$. Apply $n x^{n-1}$.

Write the coefficient as $\dfrac{1}{3}$ and the new power as $\dfrac{1}{3} - 1 = -\dfrac{2}{3}$.

The power is negative, so the derivative can be written as a fraction.

First rewrite the expression as $3x^{-2}$. Then differentiate term by term.

The coefficient is $3 \times (-2) = -6$, and the new power is $-2 - 1 = -3$.

The derivative is negative for every $x \neq 0$, so the curve falls in each of its two branches.

Differentiate each term with its own standard result. The constant in the exponent, and the constant inside the sine, both come to the front.

For $e^{2x}$ the factor is $2$, giving $8e^{2x}$. For $\sin 3x$ the factor is $3$, giving $3\cos 3x$.

The derivative of $\cos 4x$ carries a minus sign. The derivative of $\ln x$ is $\dfrac{1}{x}$, and the $2$ multiplies it.

The first term gives $-4\sin 4x$. The second gives $-\dfrac{2}{x}$.

Both terms are negative here, but for different reasons: one from the cosine rule, and one from the minus sign already in the expression.

Compare $e^{5x}$ with $e^x$. What extra factor appears when the exponent is $5x$ instead of $x$?

The derivative keeps the same exponential but gains a factor equal to the constant in the exponent.

The student forgot the constant in the exponent. The correct derivative is

A quick check: the graph of $e^{5x}$ climbs five times as fast as the graph of $e^x$, so its gradient is five times as large at every $x$.