Question #5d128

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

(1) #2^(cos(t))*ln(2)sin(t)-csin(t)(cos(t))^(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(t)#.

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) #3x^(3-1)#, which becomes #3x^2#.

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

(3) #c(cos(t))^(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(t)# is Equation 3 multiplied by #-sin(t)#, which gives

(4) #-csin(t)(cos(t))^(c-1)#.

Finally, with the first part #-2^cos(t)#, 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(t)=-e^(ln2cos(t))#

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:

#-d/dt[2^cos(t)]=-d/dt[e^(ln2cos(t))]#
#=-e^(ln2cos(t))(d/dt[ln2cos(t)])#
#=-e^(ln2cos(t))*-ln2sin(t)#
#=2^(cos(t))*ln2sin(t)#

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.