# Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. G(x) =...

## Question:

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

{eq}G(x) = \int_{x}^{2} \cos( \sqrt{7t} ) \, \mathrm{d}t {/eq}

## How to Find the Derivative Using the Fundamental Theorem of Calculus:

Finding the derivative of a function is a simple process, and when we calculate the derivative of a function using the Fundamental Theorem of Calculation it is even easier because this theorem considers the derivative as an inverse operation between derivatives and integrals. Part 1 of the Fundamental Theorem of Calculus is:

{eq}\eqalign{ & G(x) = \int_a^x {f(t)dt} \cr & \dot G(x) = \frac{d}{{dx}}\int_{a(x)}^{b(x)} {f(t)dt} \cr & \dot G(x) = f(b(x))*\dot b(x) - f(a(x))*\dot a(x) \cr} {/eq}

{eq}\eqalign{ & {\text{We find the derivative using the Part 1 of the Fundamental Theorem of Calculus}}{\text{. Given }}G(x): \cr & G(x) = \int_x^2 {\cos (\sqrt {7t} )dt} \cr & {\text{ The Theorem is:}} \cr & G(x) = \int_a^x {f(t)dt} \cr & \dot G(x) = \frac{d}{{dx}}\int_{a(x)}^{b(x)} {f(t)dt} \cr & \dot G(x) = f(b(x))*\dot b(x) - f(a(x))*\dot a(x) \cr & {\text{We substitute on equation and solve:}} \cr & \dot G(x) = \cos \left( {\sqrt {7(2)} } \right)*\dot 2 - \cos \left( {\sqrt {7(x)} } \right)*\dot x \cr & \dot G(x) = \cos \left( {\sqrt {14} } \right)*0 - \cos \left( {\sqrt {7(x)} } \right)*1 \cr & \boxed{\dot G(x) = - \cos \left( {\sqrt {7x} } \right)} \cr} {/eq} 