# Solve g(t) = \int^7_{e^{4t}} \ cos (x) \ dx .\\Calculate \ g' (t) =

## Question:

Solve {eq}g(t) = \int^7_{e^{4t}} \ cos (x) \ dx .\\Calculate \ g' (t) = {/eq}

## Calculus

{eq}\displaystyle{\text{a) Method of calculating definite integration of function is mentioned below: }\\[10pt] \int_a^b f(x)dx=\left[F(x)\right]_a^b\\[10pt] \text{where F(x) is integral value of f(x).}\\[10pt] b)\ \int \cos xdx=\sin x+C\hspace{90pt}\text{(equation 1)}\\[15pt] c)\ f(x)=\sin g(x)\\[10pt] f'(x)=\cos g(x)\times g'(x)\hspace{90pt}\text{(equation 2)}}\\ {/eq}

## Answer and Explanation:

{eq}\displaystyle{g(t) = \int^7_{e^{4t}} \ cos (x) \ dx\\[10pt] \text{Using method of definite integration}\\[10pt] g(t)=\left[\sin x\right]_{e^{4t}}^7\hspace{90pt}\text{(using equation 1)}\\[15pt] \color{blue}{g(t)=(\sin 7-\sin e^{4t})}\\[10pt] \text{Now differentiating both side with respect to t}\\[10pt] g'(t)=0-\cos e^{4t}\times e^{4t}\times 4\hspace{90pt}\text{(using equation 2) & (derivative of constant is 0 )}\\[15pt] \color{blue}{g'(t)=-4e^{4t}\cos e^{4t}}} {/eq}

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