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

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

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

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

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(e^{c o s (x)})^{^{'}}

- e^{c o s (x)} sin (x)

Solution steps

(e^{c o s (x)})^{'}

Apply the chain rule:

e^{c o s (x)} (cos (x))^{'}

= e^{c o s (x)} (cos (x))^{'}

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

= e^{c o s (x)} (- sin (x))

Simplify

= - e^{c o s (x)} sin (x)

y = 3 tan^{2} (7 x)

ln 4 + t - t^{3}

\frac{\partial}{\partial x} (x^{3} - 3 x + y^{3} - 3 y)

\int \frac{3 x + 7}{x ^{2} + 5 x + 6} d x

\int (3 x - 8)^{21} d x

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