find the derivative.\n\\(\\frac{d}{dx}\\int_{1}^{\\sin x}13t^{12}dt\\)\na. by evaluating the integral and…

find the derivative.\n\\(\\frac{d}{dx}\\int_{1}^{\\sin x}13t^{12}dt\\)\na. by evaluating the integral and differentiating the result.\nb. by differentiating the integral directly.\n\na. evaluate the definite integral.\n\\(\\frac{d}{dx}\\int_{1}^{\\sin x}13t^{12}dt = \\frac{d}{dx}(\\square)\\)\n(simplify your answer. use integers or fractions for any numbers in the expression.)

find the derivative.\n\\(\\frac{d}{dx}\\int_{1}^{\\sin x}13t^{12}dt\\)\na. by evaluating the integral and differentiating the result.\nb. by differentiating the integral directly.\n\na. evaluate the definite integral.\n\\(\\frac{d}{dx}\\int_{1}^{\\sin x}13t^{12}dt = \\frac{d}{dx}(\\square)\\)\n(simplify your answer. use integers or fractions for any numbers in the expression.)

Answer

Explanation:

Step1: Evaluate the integral

First, find the antiderivative of $13t^{12}$. The antiderivative of $t^n$ is $\frac{t^{n + 1}}{n+1}+C$, so the antiderivative of $13t^{12}$ is $13\times\frac{t^{13}}{13}=t^{13}+C$. Then, evaluate the definite - integral $\int_{1}^{\sin x}13t^{12}dt=\left[t^{13}\right]_{1}^{\sin x}=\sin^{13}x - 1$.

Step2: Differentiate the result

Differentiate $\sin^{13}x-1$ with respect to $x$. Using the chain - rule, if $y = u^{13}$ and $u=\sin x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=13u^{12}$ and $\frac{du}{dx}=\cos x$. So $\frac{d}{dx}(\sin^{13}x - 1)=13\sin^{12}x\cos x$.

Step3: Differentiate the integral directly

By the fundamental theorem of calculus and the chain - rule, if $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F^\prime(x)=f(u(x))\cdot u^\prime(x)$. Here, $a = 1$, $u(x)=\sin x$, and $f(t)=13t^{12}$. So $\frac{d}{dx}\int_{1}^{\sin x}13t^{12}dt=13(\sin x)^{12}\cdot\cos x$.

Answer:

$13\sin^{12}x\cos x$