Question

How do you find the derivative of ${\sin ^2}(2x)$ ?

Answer 1

Hint : Here, we find the first order of the derivative and by using the chain rule of the composite function. First of all we will apply the derivative of the given power in the function and then by using the chain rule will take the derivative of the function and then the derivative of the angle.

Complete step-by-step answer :
Take the given expression –
$f(x) = {\sin ^2}(2x)$
Take derivative of the given function –
$f'(x) = \dfrac{d}{{dx}}{\sin ^2}(2x)$
Apply the identity here first of all for ${x^n}$ formula in the above equation and then apply $\left( {\dfrac{u}{v}} \right)$ rule.
$f'(x) = 2.{\sin ^{2 - 1}}(2x)\dfrac{d}{{dx}}\sin (2x)$
Simplify the above equation-
$f'(x) = 2.{\sin ^1}(2x)\dfrac{d}{{dx}}\sin (2x)$
Now find the derivative of the sine function –
$f'(x) = 2.{\sin ^1}(2x)\cos (2x)\dfrac{{d(2x)}}{{dx}}$
Now, find the derivative with respect to $2x$
$f'(x) = 2.{\sin ^1}(2x)\cos (2x)2\dfrac{{d(x)}}{{dx}}$
Simplify the above equation –
$f'(x) = 2.{\sin ^1}(2x)\cos (2x)2$
The above equation can be re-written as –
$f'(x) = 4\sin (2x)\cos (2x)$
This is the required solution.
So, the correct answer is “ $f'(x) = 4\sin (2x)\cos (2x)$ ”.

Note : Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other.
In mathematics, integration is the concept of calculus and it is the act of finding the integrals. One can find the instantaneous rate of change of the function at a point by finding the derivative of that function and placing it in the x-value of the point. Instantaneous rate of change of the function can be represented by the slope of the line, which says how much the function is increasing or decreasing as the x-values change.