
How do you differentiate \[f\left( x \right)=2x\sin x\]?
Answer
556.8k+ views
Hint: Assume the given function as \[f\left( x \right)\]. Consider \[f\left( x \right)\] as the product of an algebraic function and a trigonometric function. Now, apply the product of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]. Here, consider, u = x and \[v=\sin x\]. Use the formula: - \[\dfrac{d\sin x}{dx}=\cos x\] to simplify the derivative and get the answer.
Complete step by step answer:
Here, we have been provided with the function \[2x\sin x\] and we are asked to differentiate it. Let us assume the given function as \[f\left( x \right)\]. So, we have,
\[\Rightarrow f\left( x \right)=2x\sin x\]
Now, we can assume the given function as the product of an algebraic function (x) and a trigonometric function \[\left( \sin x \right)\], which are multiplied to a constant 2. So, we have,
\[\Rightarrow f\left( x \right)=2\times x\times \sin x\]
Let us assume x as ‘u’ and \[\sin x\] as ‘v’. So, we have,
\[\Rightarrow f\left( x \right)=2\times u\times v\]
Differentiating both the sides with respect to x, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\left[ 2\left( u\times v \right) \right]}{dx}\]
Since, 2 is a constant so it can be taken out of the derivative, so we get,
\[\Rightarrow f'\left( x \right)=2\dfrac{d\left( u\times v \right)}{dx}\]
Now, applying the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow f'\left( x \right)=2\left[ u\dfrac{dv}{dx}+v\dfrac{du}{dx} \right]\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow f'\left( x \right)=2\left[ x\dfrac{d\sin x}{dx}+\sin x\dfrac{dx}{dx} \right]\]
We know that \[\dfrac{d\sin x}{dx}=\cos x\], so we have,
\[\begin{align}
& \Rightarrow f'\left( x \right)=2\left[ x\cos x+\sin x\times 1 \right] \\
& \Rightarrow f'\left( x \right)=2\left( x\cos x+\sin x \right) \\
\end{align}\]
Hence, the above relation is our answer.
Note: One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they make our question easy to solve.
Complete step by step answer:
Here, we have been provided with the function \[2x\sin x\] and we are asked to differentiate it. Let us assume the given function as \[f\left( x \right)\]. So, we have,
\[\Rightarrow f\left( x \right)=2x\sin x\]
Now, we can assume the given function as the product of an algebraic function (x) and a trigonometric function \[\left( \sin x \right)\], which are multiplied to a constant 2. So, we have,
\[\Rightarrow f\left( x \right)=2\times x\times \sin x\]
Let us assume x as ‘u’ and \[\sin x\] as ‘v’. So, we have,
\[\Rightarrow f\left( x \right)=2\times u\times v\]
Differentiating both the sides with respect to x, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\left[ 2\left( u\times v \right) \right]}{dx}\]
Since, 2 is a constant so it can be taken out of the derivative, so we get,
\[\Rightarrow f'\left( x \right)=2\dfrac{d\left( u\times v \right)}{dx}\]
Now, applying the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow f'\left( x \right)=2\left[ u\dfrac{dv}{dx}+v\dfrac{du}{dx} \right]\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow f'\left( x \right)=2\left[ x\dfrac{d\sin x}{dx}+\sin x\dfrac{dx}{dx} \right]\]
We know that \[\dfrac{d\sin x}{dx}=\cos x\], so we have,
\[\begin{align}
& \Rightarrow f'\left( x \right)=2\left[ x\cos x+\sin x\times 1 \right] \\
& \Rightarrow f'\left( x \right)=2\left( x\cos x+\sin x \right) \\
\end{align}\]
Hence, the above relation is our answer.
Note: One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they make our question easy to solve.
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