I suggest you calculate the first 2 or 3 derivatives, and see whether you can find a pattern.
derivative (7cosx) = -ln(7) 7cosx sinx dx
(cosx)^2-(sinx)^2
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
-sinx
Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
f(x) sinx f`(x) = cosx
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
The derivative is 1/(1 + cosx)
d/dx (-cscx-sinx)=cscxcotx-cosx
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).