
How do you differentiate $\sin x.\cos x$?
Answer
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Hint: We are given a product of two different trigonometric functions to be differentiated. Therefore, we must have prior knowledge of the derivatives of sine and cosine of a function. The derivatives of sine and cosine functions are different, therefore the both the terms cannot be treated as one. Hence, we shall differentiate the given function using the product rule of differentiation.
Complete step by step answer:
We use the product rule of differentiation where more than one variable is present in one term to be differentiated. We first differentiate one variable and keep the other ones unchanged, then we add another term in which we keep the first variable unchanged and differentiate the other variable.
Applying this rule, we get
$\dfrac{d}{dx}\left( \sin x.\cos x \right)=\left( \dfrac{d}{dx}\sin x \right)\cos x+\sin x.\dfrac{d}{dx}\cos x$
We know that the derivative of $\sin x$ with respect to x is $\cos x$ and the derivative of $\cos x$ with respect to x is $-\sin x$. Hence, substituting these values, we get
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)=\left( \cos x \right)\cos x+\sin x\left( -\sin x \right)$
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)={{\cos }^{2}}x+\left( -{{\sin }^{2}}x \right)$
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)={{\cos }^{2}}x-{{\sin }^{2}}x$
From the basic trigonometric identities, we also know that $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $. Since, we also have similar terms, therefore, we shall substitute the values as:
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)=\cos 2x$
Therefore, the derivative of $\sin x.\cos x$ is equal to $\cos 2x$.
Note: Another method of solving this problem was by using the basic trigonometric identity of twice the angle of sine function. It is given as $\sin 2x=2\sin x\cos x$. We would have explicitly multiplied and divided the given term by 2. Then we could have applied this property and differentiated the term $\dfrac{\sin 2x}{2}$ by using the chain rule of differentiation to first differentiate the sine function and then the angle given.
Complete step by step answer:
We use the product rule of differentiation where more than one variable is present in one term to be differentiated. We first differentiate one variable and keep the other ones unchanged, then we add another term in which we keep the first variable unchanged and differentiate the other variable.
Applying this rule, we get
$\dfrac{d}{dx}\left( \sin x.\cos x \right)=\left( \dfrac{d}{dx}\sin x \right)\cos x+\sin x.\dfrac{d}{dx}\cos x$
We know that the derivative of $\sin x$ with respect to x is $\cos x$ and the derivative of $\cos x$ with respect to x is $-\sin x$. Hence, substituting these values, we get
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)=\left( \cos x \right)\cos x+\sin x\left( -\sin x \right)$
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)={{\cos }^{2}}x+\left( -{{\sin }^{2}}x \right)$
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)={{\cos }^{2}}x-{{\sin }^{2}}x$
From the basic trigonometric identities, we also know that $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $. Since, we also have similar terms, therefore, we shall substitute the values as:
$\Rightarrow \dfrac{d}{dx}\left( \sin x.\cos x \right)=\cos 2x$
Therefore, the derivative of $\sin x.\cos x$ is equal to $\cos 2x$.
Note: Another method of solving this problem was by using the basic trigonometric identity of twice the angle of sine function. It is given as $\sin 2x=2\sin x\cos x$. We would have explicitly multiplied and divided the given term by 2. Then we could have applied this property and differentiated the term $\dfrac{\sin 2x}{2}$ by using the chain rule of differentiation to first differentiate the sine function and then the angle given.
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