
The derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$ is
A) ${{\tan }^{2}}x$
B) $\tan x$
C) $-\tan x$
D) None of these
Answer
564.6k+ views
Hint:
Here we have to find the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$. For that, we will find the differentiation of ${{\sin }^{2}}x$ with respect to x using the chain rule of differentiation by first differentiating ${{\sin }^{2}}x$ and then multiply it with the differentiation of$\sin x$. Similarly, we will find the differentiation of ${{\cos }^{2}}x$ with respect to x using the chain rule of differentiation by first differentiating ${{\cos }^{2}}x$ and multiply it with the differentiation of $\cos x$. Then we will divide the value of the derivative of ${{\sin }^{2}}x$ obtained by the value of the derivative of ${{\cos }^{2}}x$. From there, we will get the result of the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$.
Complete step by step solution:
Let ${{\sin }^{2}}x=u$and ${{\cos }^{2}}x=v$
Now, we will first differentiate u with respect to x.
$\Rightarrow \dfrac{du}{dx}=\dfrac{d{{\sin }^{2}}x}{dx}$
Here, we will use the chain rule of differentiation. We will first find the derivative of ${{\sin }^{2}}x$ and then multiply it with the differentiation of$\sin x$
Therefore,
$\Rightarrow \dfrac{du}{dx}=2\sin x.\cos x=\sin 2x\text{ }\left\{ \because \sin 2x=2\sin x.\cos x \right\}$…………….. $\left( 1 \right)$
Now, we will differentiate v with respect to x.
$\Rightarrow \dfrac{dv}{dx}=\dfrac{d{{\cos }^{2}}x}{dx}$
Here, we will use the chain rule of differentiation. We will first find the derivative of ${{\cos }^{2}}x$ and then multiply it with the differentiation of$\cos x$
Therefore,
$\Rightarrow \dfrac{dv}{dx}=2\cos x\times \left( -\sin x \right)=-\sin 2x\text{ }\left\{ \because \sin 2x=2\sin x.\cos x \right\}$ …………….. $\left( 2 \right)$
According to the question, we have to find the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$
For that, we will divide equation $\left( 1 \right)$by equation $\left( 2 \right)$
Therefore,
$\Rightarrow \dfrac{du}{dv}=\dfrac{\sin 2x}{-\sin 2x}$
On dividing numerator by denominator, we get
$\Rightarrow \dfrac{du}{dv}=-1$
We have found $\dfrac{du}{dv}$which is equal to derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$
Therefore, the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$is -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 $f(g(x))$ is equal to $f'(g(x)).g'(x)$
Here we have to find the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$. For that, we will find the differentiation of ${{\sin }^{2}}x$ with respect to x using the chain rule of differentiation by first differentiating ${{\sin }^{2}}x$ and then multiply it with the differentiation of$\sin x$. Similarly, we will find the differentiation of ${{\cos }^{2}}x$ with respect to x using the chain rule of differentiation by first differentiating ${{\cos }^{2}}x$ and multiply it with the differentiation of $\cos x$. Then we will divide the value of the derivative of ${{\sin }^{2}}x$ obtained by the value of the derivative of ${{\cos }^{2}}x$. From there, we will get the result of the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$.
Complete step by step solution:
Let ${{\sin }^{2}}x=u$and ${{\cos }^{2}}x=v$
Now, we will first differentiate u with respect to x.
$\Rightarrow \dfrac{du}{dx}=\dfrac{d{{\sin }^{2}}x}{dx}$
Here, we will use the chain rule of differentiation. We will first find the derivative of ${{\sin }^{2}}x$ and then multiply it with the differentiation of$\sin x$
Therefore,
$\Rightarrow \dfrac{du}{dx}=2\sin x.\cos x=\sin 2x\text{ }\left\{ \because \sin 2x=2\sin x.\cos x \right\}$…………….. $\left( 1 \right)$
Now, we will differentiate v with respect to x.
$\Rightarrow \dfrac{dv}{dx}=\dfrac{d{{\cos }^{2}}x}{dx}$
Here, we will use the chain rule of differentiation. We will first find the derivative of ${{\cos }^{2}}x$ and then multiply it with the differentiation of$\cos x$
Therefore,
$\Rightarrow \dfrac{dv}{dx}=2\cos x\times \left( -\sin x \right)=-\sin 2x\text{ }\left\{ \because \sin 2x=2\sin x.\cos x \right\}$ …………….. $\left( 2 \right)$
According to the question, we have to find the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$
For that, we will divide equation $\left( 1 \right)$by equation $\left( 2 \right)$
Therefore,
$\Rightarrow \dfrac{du}{dv}=\dfrac{\sin 2x}{-\sin 2x}$
On dividing numerator by denominator, we get
$\Rightarrow \dfrac{du}{dv}=-1$
We have found $\dfrac{du}{dv}$which is equal to derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$
Therefore, the derivative of ${{\sin }^{2}}x$ with respect to ${{\cos }^{2}}x$is -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 $f(g(x))$ is equal to $f'(g(x)).g'(x)$
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

