if $f(y)=e^{-3sin y+cos y}$, find $f(y)$. use exact values.\n$f(y)=$

Explanation:

Step1: Identify outer - inner functions

Let $u=-3\sin y+\cos y$, then $f(y) = e^{u}$.

Step2: Differentiate outer function

The derivative of $e^{u}$ with respect to $u$ is $e^{u}$.

Step3: Differentiate inner function

The derivative of $u=-3\sin y+\cos y$ with respect to $y$ is $u'=-3\cos y-\sin y$.

Step4: Apply chain - rule

By the chain - rule $\frac{df}{dy}=\frac{df}{du}\cdot\frac{du}{dy}$, so $f'(y)=e^{u}\cdot(-3\cos y - \sin y)$.

Step5: Substitute $u$ back

Substitute $u=-3\sin y+\cos y$ back into the expression, we get $f'(y)=e^{-3\sin y+\cos y}(-3\cos y-\sin y)$.

Answer:

$e^{-3\sin y+\cos y}(-3\cos y-\sin y)$