How do you use definition of derivatives to solve the derivative of f(x)=–2(sin(x)^5)?

1 Answer
May 5, 2015

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.