Question

Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = cos x (3 + v5)8 dv sin x

Answer 1

It looks like

$y(x)=\displaystyle\int_{\cos x}^{\sin x}(3+v^5)^8\,\mathrm dv$

(If the limits are in the wrong order, just multiply the result by -1)

Split the integral at an arbitrary value between $\cos x$ and $\sin x$, and write $y(x)$ as

$y(x)=\displaystyle\left\{\int_{\cos x}^c+\int_c^{\sin x}\right\}(3+v^5)^8\,\mathrm dv$

$y(x)=\displaystyle\int_c^{\sin x}(3+v^5)^8\,\mathrm dv-\int_c^{\cos x}(3+v^5)^8\,\mathrm dv$

Then by the FTC,

$\dfrac{\mathrm dy}{\mathrm dx}=(3+\sin^5x)^8\cdot\dfrac{\mathrm d\sin x}{\mathrm dx}-(3+\cos^5x)^8\cdot\dfrac{\mathrm d\cos x}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=\cos x(3+\sin^5x)^8+\sin x(3+\cos^5x)^8$