Solve for the derivative using:
#f'(x)=lim_{\Delta x\rarr0}{f(x+Deltax)-f(x)}/{Delta x}#
Substitute #f(x)=-2sin^5(x)#.
#f'(x)=-2lim_{Deltax\rarr0}{sin^5(x+Deltax)-sin^5(x)}/{Delta x}#
Use the sum of angles trig identity to get
#sin(x+Deltax)=sin(x)cos(\Deltax)+sin(Delta x)cos(x)#
For small angles, we have the following first order approximations
#sin(Delta x)approxDelta x# and #cos(Delta x)\approx1#
The above approximations are derived from the Taylor series expansions of sine and cosine, Since we are taking the #lim_{\Delta x rarr 0}# these approximations will become exact.
Using these approximations,
#sin(x+Deltax)\approxsin(x)+Delta x cos(x)#
Substitute this into the #f'(x)# expression,
#f'(x)=-2lim_{Deltaxrarr0}{[sin(x)+Deltaxcos(x)]^5-sin^5(x)}/{Delta x}#
Since we are taking #lim_{Delta xrarr0}#
#(Delta x)^2# will be soooooo much smaller than #Delta x# to the point where I can write:
#[sin(x)+Delta x cos(x)]^5=sin^5(x)+5Delta x cos(x)sin^4(x)#
and ignore all the other terms with #Delta x# raised to a higher power. If you don't mind all the extra writing, you can write out all of the terms and still get the right answer in the end.
Substitute the above expression into #f'(x)#
#f'(x)=-2\lim_{Deltaxrarr0}{[sin^5(x)+5Delta x cos(x)sin^4(x)]-sin^5(x)}/{Delta x}#
#f'(x)=-2\lim_{Deltaxrarr0}{5Delta x cos(x)sin^4(x)}/{Delta x}#
#f'(x)=-10cos(x)sin^4(x)#
The expected answer.
