Question

What is the derivative of $\sin x(\sin x+\cos x)$?

Answer 1

Hint: We use Product – Rule to find the derivative of function $\sin x(\sin x+\cos x)$.
The product rule helps us to differentiate between two or more of the functions in a given function.
If $u$ and $v$ are the two given function of $x$ then the product rule is given by the following formula:
${\displaystyle \frac{d(uv)}{dx}}=u{\displaystyle \frac{dv}{dx}}+v{\displaystyle \frac{du}{dx}}$. After product rule we use different trigonometric identities such as $\cos 2x={\cos }^{2}x-{\sin }^{2}x$ and $\sin 2x=2\sin x\cos x$ to solve the given problem.

Complete step by step answer:
The given function is the product of two functions $\sin x$ and $(\sin x+\cos x)$.
We use product rule to find the derivative, that means first we multiply the first function by the derivative of the second function and the second function is multiplied by the derivative of the first function and add them.
So, we can write it as
$\Rightarrow \sin x\left({\displaystyle \frac{d(\sin x+\cos x)}{dx}}\right)+(\sin x+\cos x)\left({\displaystyle \frac{d(\sin x)}{dx}}\right)\dots \dots \dots (1)$
We know that the derivative of $\sin x=\cos x$ and the derivative of $\cos x=-\sin x$.
So, ${\displaystyle \frac{d(\sin x+\cos x)}{dx}}=(\cos x-\sin x)$ and ${\displaystyle \frac{d\sin x}{dx}}=\cos x$
Put these values in equation $1$. We get,
$\Rightarrow \sin x(\cos x-\sin x)+(\sin x+\cos x)(\cos x)$
Simplifying the above equation. We get,
$\Rightarrow \sin x\cos x-{\sin }^{2}x+\sin x\cos x+{\cos }^{2}x$
$\Rightarrow 2\sin x\cos x+{\cos }^{2}x-{\sin }^{2}x$
We know that $\cos 2x={\cos }^{2}x-{\sin }^{2}x$ and $\sin 2x=2\sin x\cos x$.
Write these values in the above equation. We get,
$\Rightarrow \sin 2x+\cos 2x$
Hence, the derivative of the function $\sin x(\sin x+\cos x)$ is $\sin 2x+\cos 2x$.

Note:
To find the derivative of the given function we can also use chain rule. First we simplify the expression by opening brackets and we will get a composite function. Chain rule is used where the function is composite, we can denote chain rule by $f.g$, where $f$ and $g$ are two functions. Chain rule states that the derivative of a composite function can be taken as the derivative of the outer function which we multiply by the derivative of the inner function.