Question

Find: dy/dx, if y=x sinx + cos x

Answer 1

Differentiation

We are asked to find the derivative of the following equation:

$\longrightarrow y = x \sin(x) + \cos(x)$

We can solve this problem using product rule. In calculus, the product rule, also called Leibniz's law, governs the differentiation of products of differentiable functions. It may be stated as:

$\;{(f \cdot g)' = f' \cdot g + f \cdot g'}$

We will be using the following identities to solve the problem.

$\boxed{\begin{array}{l} \textsf{\textbf{IDENTITIES}} \\ \\ \bull\:\:\dfrac{d}{dx}(x) = 1 \\ \\ \bull\:\:\dfrac{d}{dx}(\sin(x)) = \cos(x) \\ \\ \bull\:\:\dfrac{d}{dx}(\cos(x)) = -\sin(x) \\ \\ \bull\:\: (f \cdot g)' = f' \cdot g + f \cdot g'\end{array}}$

Now, consider,

$\implies \dfrac{d}{dx}(x \sin(x) + \cos(x))$

By using the product in the above equation, we get the following results:

$\implies \sin(x) \cdot \dfrac{d}{dx}(x) + x \cdot \dfrac{d}{dx}(\sin(x)) + \dfrac{d}{dx}(\cos(x))$

$\implies \sin(x) \cdot \dfrac{d}{dx}(x) + x \cdot \dfrac{d}{dx}(\sin(x)) + \dfrac{d}{dx}(\cos(x))$

$\implies \sin(x) \cdot 1 + x \cdot \cos(x) + (-(\sin(x)))$

$\implies \sin(x) \cdot 1 + x \cdot \cos(x) - \sin(x))$

$\implies \boxed{x \cdot \cos(x)}$

Hence, this is our essential solution.