Derivative of Trigonometric Functions

Find the derivative of \( f(x)=\sin{2x}.\)

Answer:

To find this derivative you will need to use the Chain Rule. Let \( u=2x.\) Then by the Power Rule,

$$u'(x)=2.$$

So now using the Chain Rule,

$$\begin{align}f'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\sin{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left( \cos{u} \right) (2) \\ &= 2\cos{u}. \end{align}$$

Finally, substitute back \( u=2x, \) so

$$f'(x)=2\cos{2x}.$$

Find the derivative of \( g(x)=\tan{x^3}.\)

Answer:

Start by letting \( u=x^3.\) By the Power Rule,

$$u'(x)=3x^2.$$

Next, use the Chain Rule,

$$\begin{align}g'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\tan{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(\sec^2{u} \right) (3x^2) \\ &= 3x^2\sec^2{u}, \end{align}$$

and substitute back \( u=x^3,\) obtaining

$$g'(x)=3x^2\sec^2{x^3}.$$

Find the derivative of \( h(x)=\csc{2x^2}.\)

Answer:

Start by letting \( u=2x^2.\) By the Power Rule,

$$u'(x)=4x.$$

Next, use the Chain Rule to get

$$\begin{align}h'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\csc{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(-\csc{u} \right) \left( \cot{u} \right) (4x) \\ &= -4x\left(\csc{u}\right) \left(\cot{u} \right) , \end{align}$$

and substitute back \( u=2x^2\) to get

$$h'(x)=-4x\left(\csc{2x^2}\right) \left(\cot{2x^2} \right).$$

Find the derivative of \( f(x)=x \left(\sin{x}\right).\)

Answer:

Since you have a product of functions start by using the Product Rule, that is

$$f'(x)=\left( \frac{\mathrm{d}}{\mathrm{d}x} x \right) \sin{x} + x \left(\frac{\mathrm{d}}{\mathrm{d}x} \sin{x} \right).$$

You can find the derivative of \( x \) by using the Power Rule,

$$\frac{\mathrm{d}}{\mathrm{d}x} x= 1,$$

and the derivative of the sine function is the cosine function

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

Knowing this, the derivative of \( f(x) \) is

$$\begin{align}f'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}x} x \right) \sin{x} + x \left(\frac{\mathrm{d}}{\mathrm{d}x} \sin{x} \right) \\[0.5em] &= (1)\sin{x}+x\left( \cos{x} \right) \\ &= \sin{x}+x \cos{x}.\end{align}$$

Find the derivative of $$ g(x) = \frac{\tan{x}}{x^2}.$$

Answer:

Now you have a quotient of functions, so start by using the Quotient Rule, that is

$$g'(x)=\frac{ \left( \frac{\mathrm{d}}{\mathrm{d}x} \tan{x} \right)x^2-\tan{x}\left( \frac{\mathrm{d}}{\mathrm{d}x} x^2 \right) }{\left( x^2 \right)^2}.$$

You can find the derivative of \( x^2 \) with the Power Rule,

$$\frac{\mathrm{d}}{\mathrm{d}x} x^2=2x,$$

and the derivative of the tangent function is the secant function squared

$$\frac{\mathrm{d}}{\mathrm{d}x} \tan{x}=\sec^2{x}.$$

Finally, substitute the above derivatives in the Quotient Rule and simplify, obtaining

$$\begin{align}g'(x) &= \frac{ \left( \sec^2{x} \right) x^2- \left( \tan{x} \right) (2x) }{ \left( x^2 \right) ^2} \\[0.5em] &= \frac{x^2 \left( \sec^2{x} \right) -2x \left( \tan{x} \right) }{x^4} \\[0.5em] &= \frac{x \left( \sec^2{x} \right) - 2\tan{x}}{x^3} .\end{align}$$

Find the derivative of \( h(x)=\sin^2{x}.\)

Answer:

Since the sine function is squared, you are dealing with a composition of functions, hence you need to use the Chain Rule. Start by letting \( u=\sin{x}.\) Its, derivative is the cosine function

$$u'(x)=\cos{x}.$$

Next, use the Chain Rule,

$$\begin{align}h'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}u^2 \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(2u \right) \left( \cos{x} \right) \\ &= 2u \left(\cos{u} \right), \end{align}$$

where you have used the Power Rule to find the derivative of \( u^2.\) Finally, substitute back \( u=\sin{x},\) obtaining

$$h'(x)=2 \left( \sin{x} \right) \left( \cos{x} \right) .$$

One common mistake is getting the signs mistaken when differentiating the cosine function, the cotangent function, or the cosecant function, that is

$$\frac{\mathrm{d}}{\mathrm{d}x} \cos{x} \neq \sin{x}. $$

Be sure to put the negative sign!

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

Another common mistake happens when differentiating the secant function or the cosecant function. Remember that when differentiating these functions you have to write the correct inputs in all instances of trigonometric functions.

$$\frac{\mathrm{d}}{\mathrm{d}x} \sec{x^2} \neq 2x \left(\sec{x^2} \right) \left( \tan{x} \right). $$

Here, the square of the input of the tangent function is missing. Find the derivative using the derivative of the secant function formula. Do not forget to use any relevant differentiation techniques, like the Chain Rule in this case.

$$\frac{\mathrm{d}}{\mathrm{d}x} \sec{x^2} = 2x \left(\sec{x^2} \right) \left( \tan{x^2} \right). $$