What is the derivative of sin^2(x) + sinx?

Jul 29, 2015

${y}^{'} = \cos x \cdot \left(1 + 2 \sin x\right)$

Explanation:

This derivative can be found by using the chain rule, the power rule, and the sum rule.

Notice that you can write your function as the sum of two other functions, let's say $f \left(x\right)$ and $g \left(x\right)$, so that you have

$y = {\underbrace{{\sin}^{2} x}}_{\textcolor{b l u e}{f \left(x\right)}} + {\overbrace{\sin x}}^{\textcolor{p u r p \le}{g \left(x\right)}}$

The sum rule tells you that you can differentiate the sum of two functions as the sum of the derivatives of those two functions

color(blue)(d/dx(y) = f^'(x) + g^'(x)

In your case, you can write

${y}^{'} = \frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) + \frac{d}{\mathrm{dx}} \left(\sin x\right)$

It's always useful to remember that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

This means that you have

${y}^{'} = \frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) + \cos x$

Now focus on the derivative of $f \left(x\right)$. In order to differentiate this function, you're going to use the chain rule, which states that the derivative of a function that depends on a variable $u$, which in turn depends on a variable $x$, can be found by

color(blue)(d/dx(y) = d/(du)(y) * d/dx(u)

In your case, you have $\sin x = u$, which implies that ${\sin}^{2} x = {u}^{2}$. The derivative will be

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = \frac{d}{\mathrm{du}} \left({u}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

Here is where the power rule comes into play. For a variable $x$ raised to a power $a$, the derivative can be found by

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left({x}^{a}\right) = a \cdot {x}^{a - 1}}$

This means that you have

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = 2 {u}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = 2 \cdot u \cdot \cos x$

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = 2 \cdot \sin x \cdot \cos x$

Your original derivative now becomes

${y}^{'} = 2 \sin x \cdot \cos x + \cos x$

which can be written as

${y}^{'} = \textcolor{g r e e n}{\cos x \cdot \left(1 + 2 \sin x\right)}$