Question

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

Answer 1

Solve the question using the first principle of derivatives. Given a function $y = f(x)$, it’s the first derivative, the rate of change of y with respect to the change in x, is defined by $\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{f(x + h) - f(x)}}{{(x + h) - (x)}}} \right]$.

Complete step by step answer:
Finding the derivative of a function by calculating the limit is known as differentiation from first principles. Use the identity $\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B = {\cos ^2}B - {\cos ^2}A$
We are asked to find the derivative of $y = {\cos ^2}(2x)$………..(1)
Increase $y$ to $y + \delta y$ and correspondingly $x$ increase to $x + \delta x$ in the above equation.
$\Rightarrow y + \delta y = {\cos ^2}\left( {x + \delta x} \right)$ ………….. (2)
Subtract equation 1 from equation 2, we get
$\Rightarrow y + \delta y - y = {\cos ^2}(x + \delta x) - {\cos ^2}x$
$\Rightarrow \delta y = {\cos ^2}(x + \delta x) - {\cos ^2}x$

Now, divide both sides of the equation with $\delta x$
$\Rightarrow \dfrac{{\delta y}}{{\delta x}} = \dfrac{{{{\cos }^2}(x + \delta x) - {{\cos }^2}x}}{{\delta x}}$
Using $\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B = {\cos ^2}B - {\cos ^2}A$
We reach to the following step,
$\Rightarrow \dfrac{{\delta y}}{{\delta x}} = - \dfrac{{\sin (2x + \delta x)\sin \delta x}}{{\delta x}}$
Take limit on both side with respect to x where x approaches to 0
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\delta x}}{{\delta y}} = \mathop {\lim }\limits_{x \to 0} - \dfrac{{\sin (2x + \delta x)\sin \delta x}}{{\delta x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{x \to 0} - \dfrac{{\sin (2x + \delta x)\sin \delta x}}{{\delta x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \sin (2x + 0) \times 1$
Because $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \delta x}}{{\delta x}} = 1$

Therefore, the answer comes out to be $\dfrac{{dy}}{{dx}} = - \sin 2x = - 2\sin x\cos x$.

Note: Differentiability of a function: A function $f(x)$ is differentiable at $x = a$ in its domain, it it’s derivative is continuous at $a$. That implies $f'(a)$ must exist , or equivalently : $\mathop {\lim }\limits_{x \to {a^ + }} f'(x) = \mathop {\lim }\limits_{x \to {a^ - }} f'(x) = \mathop {\lim }\limits_{x \to a} f'(x) = f'(a)$. A continuous function is always differentiable but a differentiable function may not be continuous. Therefore, it is advised to students that continuity checks should be performed before finding the derivative of a function.