# Question #5d128

Mar 12, 2015

Assuming $c$ is some constant, the answer is:

(1) ${2}^{\cos \left(t\right)} \cdot \ln \left(2\right) \sin \left(t\right) - c \sin \left(t\right) {\left(\cos \left(t\right)\right)}^{c - 1}$.

To do this problem, you need to focus each part of the equation to differentiate before putting them together. So, let's start with ${\cos}^{c} \left(t\right)$.

We know that according to the power rule, you would bring the power down to multiply the function, and then subtract by one on the exponent. For example, if you have ${x}^{3}$, you would bring down the 3 as follows and subtract the 3 exponent by 1:

(2) $3 {x}^{3 - 1}$, which becomes $3 {x}^{2}$.

However, in this case we only have some constant variable c, so the exponent would be

(3) $c {\left(\cos \left(t\right)\right)}^{c - 1}$.

In addition, we also need to do the chain rule, which states that you take a derivative on the inner function to multiply by the outer function. Since cos(t) is the inner function, the total derivative of ${\cos}^{c} \left(t\right)$ is Equation 3 multiplied by $- \sin \left(t\right)$, which gives

(4) $- c \sin \left(t\right) {\left(\cos \left(t\right)\right)}^{c - 1}$.

Finally, with the first part $- {2}^{\cos} \left(t\right)$, there are different ways to solve for it including implicit differentiation, but I am going to convert this function to natural e base that equals the function:

(5) $- {2}^{\cos} \left(t\right) = - {e}^{\ln 2 \cos \left(t\right)}$

This makes it easier to take a derivative, since any derivative of natural e base equals its anti-derivative. However, we need to take the chain rule to get the complete derivative and simplify, as shown in the step-by-step procedure:

$- \frac{d}{\mathrm{dt}} \left[{2}^{\cos} \left(t\right)\right] = - \frac{d}{\mathrm{dt}} \left[{e}^{\ln 2 \cos \left(t\right)}\right]$
$= - {e}^{\ln 2 \cos \left(t\right)} \left(\frac{d}{\mathrm{dt}} \left[\ln 2 \cos \left(t\right)\right]\right)$
$= - {e}^{\ln 2 \cos \left(t\right)} \cdot - \ln 2 \sin \left(t\right)$
$= {2}^{\cos \left(t\right)} \cdot \ln 2 \sin \left(t\right)$

Now that the parts we need to differentiate are complete, we can then substitute them back and get the derivative of the original function equaling Eq. 1.