What is the (e^{sin(x)})^' ?

The (e^{sin(x)})^' is e^{sin(x)}cos(x)

What is the first (e^{sin(x)})^' ?

The first (e^{sin(x)})^' is e^{sin(x)}cos(x)

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(e^{s i n (x)})^{^{'}}

e^{s i n (x)} cos (x)

Solution steps

(e^{s i n (x)})^{'}

Apply the chain rule:

e^{s i n (x)} (sin (x))^{'}

= e^{s i n (x)} (sin (x))^{'}

(sin (x))^{'} = cos (x)

= e^{s i n (x)} cos (x)

\int_{1}^{4} e^{x} d x

y = (x^{2} + 1)^{5} (1 - x^{2})^{5}

f (x) = x^{3} - 2 x^{2} + 5

f (x) = 3 x^{3} + 6 x^{2} + 7 x

\int \frac{2 x ^{4} + x ^{2} - x - 2}{x ^{3} ( x + 1 )} d x