Solveeit Logo
Login

Question

Question: The derivative of \({{\sin }^{2}}x\) with respect to \({{\cos }^{2}}x\) is A) \({{\tan }^{2}}x\) ...

The derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}x is
A) tan⁡2x{{\tan }^{2}}x
B) tan⁡x\tan x
C) −tan⁡x-\tan x
D) None of these

Explanation

Solution

Here we have to find the derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}x. For that, we will find the differentiation of sin⁡2x{{\sin }^{2}}x with respect to x using the chain rule of differentiation by first differentiating sin⁡2x{{\sin }^{2}}x and then multiply it with the differentiation ofsin⁡x\sin x. Similarly, we will find the differentiation of cos⁡2x{{\cos }^{2}}x with respect to x using the chain rule of differentiation by first differentiating cos⁡2x{{\cos }^{2}}x and multiply it with the differentiation of cos⁡x\cos x. Then we will divide the value of the derivative of sin⁡2x{{\sin }^{2}}x obtained by the value of the derivative of cos⁡2x{{\cos }^{2}}x. From there, we will get the result of the derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}x.

Complete step by step solution:
Let sin⁡2x=u{{\sin }^{2}}x=uand cos⁡2x=v{{\cos }^{2}}x=v
Now, we will first differentiate u with respect to x.
⇒dudx=dsin⁡2xdx\Rightarrow \dfrac{du}{dx}=\dfrac{d{{\sin }^{2}}x}{dx}
Here, we will use the chain rule of differentiation. We will first find the derivative of sin⁡2x{{\sin }^{2}}x and then multiply it with the differentiation ofsin⁡x\sin x
Therefore,
\Rightarrow \dfrac{du}{dx}=2\sin x.\cos x=\sin 2x\text{ }\left\\{ \because \sin 2x=2\sin x.\cos x \right\\}…………….. (1)\left( 1 \right)
Now, we will differentiate v with respect to x.
⇒dvdx=dcos⁡2xdx\Rightarrow \dfrac{dv}{dx}=\dfrac{d{{\cos }^{2}}x}{dx}
Here, we will use the chain rule of differentiation. We will first find the derivative of cos⁡2x{{\cos }^{2}}x and then multiply it with the differentiation ofcos⁡x\cos x
Therefore,
\Rightarrow \dfrac{dv}{dx}=2\cos x\times \left( -\sin x \right)=-\sin 2x\text{ }\left\\{ \because \sin 2x=2\sin x.\cos x \right\\} …………….. (2)\left( 2 \right)
According to the question, we have to find the derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}x
For that, we will divide equation (1)\left( 1 \right)by equation (2)\left( 2 \right)
Therefore,
⇒dudv=sin⁡2x−sin⁡2x\Rightarrow \dfrac{du}{dv}=\dfrac{\sin 2x}{-\sin 2x}
On dividing numerator by denominator, we get
⇒dudv=−1\Rightarrow \dfrac{du}{dv}=-1
We have found dudv\dfrac{du}{dv}which is equal to derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}x
Therefore, the derivative of sin⁡2x{{\sin }^{2}}x with respect to cos⁡2x{{\cos }^{2}}xis -1.

Hence, the correct option is D.

Note:
We need to know the following terms as we have used them in this solution.
Differentiation is defined as a process in which we find the function which gives the output of rate of change of one variable with respect to another variable.
Differentiation by chain rule: In this method, the differentiation of function f(g(x))f(g(x)) is equal to f′(g(x)).g′(x)f'(g(x)).g'(x)