Question

1. - / 1 points find the derivative of the function. y = x + (x + sin²(x))⁵⁷ y =

Explanation:

Step1: Apply chain - rule

Let $u=x+(x + \sin^{2}(x))^{5}$, so $y = u^{7}$. Then $y'=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=7u^{6}=7[x+(x + \sin^{2}(x))^{5}]^{6}$.

Step2: Find $\frac{du}{dx}$

Let $v=x+\sin^{2}(x)$. Then $u=x + v^{5}$. So $\frac{du}{dx}=1 + 5v^{4}\cdot\frac{dv}{dx}$.

Step3: Find $\frac{dv}{dx}$

$\frac{dv}{dx}=1+2\sin(x)\cos(x)$ (using the sum - rule and the chain - rule for $\sin^{2}(x)$ where if $w = \sin(x)$, then $\sin^{2}(x)=w^{2}$ and $\frac{d}{dx}(w^{2}) = 2w\cdot w'=2\sin(x)\cos(x)$).

Step4: Substitute $v$ back into $\frac{du}{dx}$

$\frac{du}{dx}=1 + 5(x+\sin^{2}(x))^{4}(1 + 2\sin(x)\cos(x))$.

Step5: Calculate $y'$

$y'=7[x+(x + \sin^{2}(x))^{5}]^{6}[1 + 5(x+\sin^{2}(x))^{4}(1 + 2\sin(x)\cos(x))]$.

Answer:

$7[x+(x + \sin^{2}(x))^{5}]^{6}[1 + 5(x+\sin^{2}(x))^{4}(1 + 2\sin(x)\cos(x))]$