Understanding Derivatives of Trigonometric Functions

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Math › Understanding Derivatives of Trigonometric Functions

Questions 1 - 4

1

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=?$

$\cos(x)$

$\sin(x)$

$\csc(x)$

$\sec(x)$

$-\sin(x)$

Explanation

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

2

What is the second derivative of $\sin(x)$ ?

$-\sin(x)$

$\cos(x)$

$-\cos(x)$

$\sec(x)$

$\csc(x)$

Explanation

The derivatives of trig functions must be memorized. The first derivative is:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

To find the second derivative, we take the derivative of our result.

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$ .

Therefore, the second derivative will be $-\sin(x)$ .

3

Find the derivative of the following function:

$f(x) = 3 ;\textup{sin}(x)$

$f(x) = 3 ;\textup{cos}(x)$

$f(x) = -3 ;\textup{cos}(x)$

$f(x) = 3 ;\textup{tan}(x)$

$f(x) = 3 ;\textup{csc}(x)$

$f(x) = -3 ;\textup{sin}(x)$

Explanation

The derivative of $\sin (x)$ is $\cos (x)$ . It is probably best to memorize this fact (the proof follows from the difference quotient definition of a derivative).

Our function

$f(x) = 3 ;\textup{sin}(x)$

the factor of 3 does not change when we differentiate, therefore the answer is

$f(x) = 3 ;\textup{cos}(x)$

4

Compute the derivative of the function $f(x) = \tan x^{2}$ .

$f' (x) = 2x \sec^{2} x^{2}$

$f' (x) = x^{2} \sec^{2} x^{2}$

$f' (x) = x^{2} \tan x^{2} + 2x \sec^{2} x^{2}$

$f' (x) = 2x \sec x^{2}$

$f' (x) = x^{2} \sec 2x$

Explanation

Use the Chain Rule.

Set $u = x^{2}$ and substitute.

$f(x) = \tan u$

$\frac{\mathrm{d} f }{\mathrm{d} u} = \frac{\mathrm{d} }{\mathrm{d} u} \tan u = \sec^{2} u$

$\frac{\mathrm{d} u }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} x^{2} = 2x^{2-1} =2x$

$\frac{\mathrm{d} f }{\mathrm{d} x} = \frac{\mathrm{d} f }{\mathrm{d} u} \cdot \frac{\mathrm{d} u }{\mathrm{d} x}= \sec^{2} u \cdot 2x = 2x \sec^{2} x^{2}$

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