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

Answer 1

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the derivative of ( f(x) = -2(\sin(x))^5 ), we can use the definition of derivatives. The derivative of a function ( f(x) ) is given by the limit:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Applying this definition to ( f(x) = -2(\sin(x))^5 ), we have:

[ f'(x) = \lim_{h \to 0} \frac{-2(\sin(x + h))^5 + 2(\sin(x))^5}{h} ]

Using the trigonometric identity ( \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) ), we can expand ( (\sin(x + h))^5 ) as follows:

[ (\sin(x + h))^5 = (\sin(x)\cos(h) + \cos(x)\sin(h))^5 ]

Expanding this using the binomial theorem, we get:

[ (\sin(x + h))^5 = \sin^5(x) + 5\sin^4(x)\cos(x)h + 10\sin^3(x)\cos^2(x)h^2 + 10\sin^2(x)\cos^3(x)h^3 + 5\sin(x)\cos^4(x)h^4 + \cos^5(x)h^5 ]

Substituting this back into the derivative expression and simplifying, we find:

[ f'(x) = \lim_{h \to 0} \frac{-2(\sin(x)\cos(h) + \cos(x)\sin(h))^5 + 2(\sin(x))^5}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{-2(\sin^5(x) + 5\sin^4(x)\cos(x)h + 10\sin^3(x)\cos^2(x)h^2 + 10\sin^2(x)\cos^3(x)h^3 + 5\sin(x)\cos^4(x)h^4 + \cos^5(x)h^5) + 2(\sin(x))^5}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{-10(\sin^4(x)\cos(x)h + \sin^3(x)\cos^2(x)h^2 + 5\sin^2(x)\cos^3(x)h^3 + \sin(x)\cos^4(x)h^4) - 2(\sin(x)\cos(h))^5 + 2(\sin(x))^5}{h} ]

[ f'(x) = -10\sin^4(x)\cos(x) - 2(\sin(x))^5 ]

Therefore, the derivative of ( f(x) = -2(\sin(x))^5 ) is ( f'(x) = -10\sin^4(x)\cos(x) ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7