# How do you sketch the graph y=sinx+sin^2x using the first and second derivatives from 0<=x<2pi?

Jun 12, 2017

#### Explanation:

We have $y = \sin x + {\sin}^{2} x = \sin x \left(1 + \sin x\right)$

Observe that $y = 0$ when $\sin x = 0$ or $1 + \sin x = 0$

hence between $0 \le x \le 2 \pi$, it intersects $x$-axis at $x = \left\{0 , \pi , \frac{3 \pi}{2} , 2 \pi\right\}$ or $\left\{0 , 3.1416 , 4.7124 , 6.2832\right\}$

Now we have extrema when $\frac{\mathrm{dy}}{\mathrm{dx}}$ and as

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x + 2 \sin x \cos x = \cos x \left(1 + 2 \sin x\right)$ or $\cos x + \sin 2 x$

also $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \sin x + 2 \cos 2 x$

Hence extrema occurs when $\cos x = 0$ i.e. $\left\{\frac{\pi}{2} , \frac{3 \pi}{2}\right\}$. While second derivative is negative at both these points, when $1 + 2 \sin x = 0$ i.e. $\sin x = - \frac{1}{2}$ or $x = \left\{\frac{5 \pi}{3} , \frac{7 \pi}{3}\right\}$, it is positive.

Hence we have a local minima at $x = \left\{\frac{5 \pi}{3} , \frac{7 \pi}{3}\right\}$ and local maxima at $\left\{\frac{\pi}{2} , \frac{3 \pi}{2}\right\}$.

The graph appears as shown below.

graph{sinx(1+sinx) [-2.917, 7.083, -1.84, 3.16]}