# Find the function f(x), whose derivative f'(x)=\cos x+ \sin x

## Question:

Find the function {eq}f(x) {/eq}, whose derivative {eq}f'(x)=\cos x+ \sin x {/eq}

## Antiderivative:

When we have the derivative of the function and we have to evaluate the original function, to evaluate this we integrate the given derivative of the function. This process is known as anti-differentiation that is the reverse of differentiation.

To evaluate the integral, we 'll apply the following rules of integration:

{eq}\int \cos xdx=\sin x+C\\ \int \sin xdx=-\cos x+C {/eq}

We have to find the function {eq}f(x) {/eq}, whose derivative {eq}f'(x)=\cos x+ \sin x {/eq}

To evaluate the given function, we integrate the given derivative of the function as differentiation and integration are the opposite of each other.

Integrate the given derivative of the function with respect to {eq}x {/eq}

{eq}\begin{align} f(x) &=\int \left (\cos x+ \sin x \right )dx\\ &=\int \cos xdx+\int \sin x \ dx \ & \left [ \int (f(x)\pm g(x))dx=\int f(x)dx \pm \int g(x)dx \right ]\\ &=\sin x-\cos x+C \ & \left [ \int \cos xdx=\sin x+C, \int \sin xdx=-\cos x+C \right ] \end{align} {/eq}

where {eq}C {/eq} is a constant of integration.