Use the limit definition to compute f′(a) and find an equation of the tangent line. f(x)=1/sqrt(64x), a=4. y=____.

Answer 1

The definition of the derivative is limit as h approaches 0 of [f(x+h)-f(x)]/h

f(x)=1/sqrt(64x) but this can be simplified by taking the square root of 64 so f(x) = 1/(8sqrt(x))

so f(x + h) = 1/(8sqrt(x+h))

That means

Get a common denominator to combine these. Multiply the left side by sqrt(x)/sqrt(x) and multiply the right side by sqrt(x+h)/sqrt(x+h)

That results in [f(x+h) - f(x)]/h = [sqrt(x) - sqrt(x+h)]/[8hsqrt(x² + xh)]

This is a complex limit. To resolve it, multiply top and bottom by the conjugate of the top [sqrt(x) + sqrt(x+h)]

This gets into a bunch of algebra, but when you get done, you are left with:

limit as h approaches 0 of -1/[8sqrt(x²+xh)(sqrt(x) + sqrt(x+h))] which then allows you to plug in 0 for h to simplify it as f'(x) = -1/(16xsqrt(x)

So f'(4) = -1/128 That is the slope of the tangent line.

To find a point on the line, we can let x = 4 in the original equation. If x = 4, then f(x) = 1/(8sqrt(4)) = 1/16 so a point on the line is (4, 1/16). The equation, using point-slope form, is y - 1/16 = -1/128(x - 4) or y = -1/128x + 1/32 + 1/16 or y = -1/128x + 3/32

Answer 2

if f(x) = 1/sqrt(64x) = (64x)^-(1/2)

Then

f'(x) = -32/(64x)^(3/2)

and

f'(4) = -32/(64*4)^(3/2) = -.0078125

@ x=4, f(4) = 1/sqrt(64*4) = .0625

So, the tangent line would have the form y-.0625 = -.0078125(x-4) or y = -.0078125x + .09375