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Question

Question: How do you differentiate \[f\left( x \right)=2x\sin x\]?...

How do you differentiate f(x)=2xsinxf\left( x \right)=2x\sin x?

Explanation

Solution

Assume the given function as f(x)f\left( x \right). Consider f(x)f\left( x \right) as the product of an algebraic function and a trigonometric function. Now, apply the product of differentiation given as: - d(u×v)dx=udvdx+vdudx\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}. Here, consider, u = x and v=sinxv=\sin x. Use the formula: - dsinxdx=cosx\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 2xsinx2x\sin x and we are asked to differentiate it. Let us assume the given function as f(x)f\left( x \right). So, we have,
f(x)=2xsinx\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 (sinx)\left( \sin x \right), which are multiplied to a constant 2. So, we have,
f(x)=2×x×sinx\Rightarrow f\left( x \right)=2\times x\times \sin x
Let us assume x as ‘u’ and sinx\sin x as ‘v’. So, we have,
f(x)=2×u×v\Rightarrow f\left( x \right)=2\times u\times v
Differentiating both the sides with respect to x, we get,
df(x)dx=d[2(u×v)]dx\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,
f(x)=2d(u×v)dx\Rightarrow f'\left( x \right)=2\dfrac{d\left( u\times v \right)}{dx}
Now, applying the product rule of differentiation given as: - d(u×v)dx=udvdx+vdudx\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}, we get,
f(x)=2[udvdx+vdudx]\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,
f(x)=2[xdsinxdx+sinxdxdx]\Rightarrow f'\left( x \right)=2\left[ x\dfrac{d\sin x}{dx}+\sin x\dfrac{dx}{dx} \right]
We know that dsinxdx=cosx\dfrac{d\sin x}{dx}=\cos x, so we have,

& \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{aligned}$$ Hence, the above relation is our answer. **Note:** One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, $$\dfrac{u}{v}$$ rule etc. as they make our question easy to solve.