Find the derivative

Question

Suppose h(x)=sqrt(f(x)) and the equation of the tangent line to f(x) at x=1 is
y=4+5(x−1).
Find h′(1).

Answer 1

h(x) = sqrt(f(x))

=> f(x) = (h(x))^2

=> f'(x) = 2 * h(x) * h'(x) = 2 * sqrt(f(x)) * h'(x)

=> h'(x) = f'(x) / [2 * sqrt(f(x))]

 

at x = 1, h'(1) = f'(1) / [2 * sqrt(f(1))]

 

the equation of the tangent line t(x) to f(x) at x=1 is: t(x) = 4 + 5(x−1) = 5 * x -1

 

since t(x) is tangent to f(x) at x = 1,

so f(x) = t(x) and f'(x) = t'(x) at x = 1

=> f(1) = t(1) = 5 * 1 - 1 = 4 and f'(1) = t'(1) = 5

h′(1) = f'(1) / [2 * sqrt(f(1))] = 5 / [2 * sqrt(4)] = 5 / 4

Answer 2

h'(x) = 0.5 * f(x)^(-1/2)*f'(x)

 

The line and the curve intercept at (1,4) ==> f(1) = 4

The slope of the curve at x=1 is the slope of the line ==> f'(1) = 5

 

h'(1) = 0.5 * 0.5 * 5 = 1.25

Answer 3

The tangent line to f(x) at  x = 1 is y(x)=5(x-1)+4 which means

y(1)=f(1)=4. the derivative of h(x) is h`(x)=f`(x)/(2sqrt(f(x))). We know that f`(1)=5 because the line tangent at that point has a slope of 5. So evidently

                        h`(1)=f`(1)/(2*sqrt(f(1)))=5/(2sqrt(4))=5/4

 

Answer 4

From the function rule of derivative, we have: h'(x) = f'(x) * 1/2 * 1/sqrt(f(x)).

 

The tangent line to f(x) at x = 1 is: y = 4 + 5 * (x - 1), so we have: f(1) = 4 + 5 * (1 - 1) = 4. The slope of the line that is tangent at x = 1 to f(x) is equal to 5, therefore, f'(1) = 5.

 

Then we have:

 

h'(1) = f'(1) * 1/2 * 1/sqrt(f(1)) = 5 * 1/2 * 1/sqrt(4) = 5/4.

Answer 5

Start by using the chain rule. h=f^1/2 so then h' is (1/2)f^-1/2*f'. 

 

Now we need to figure out what f' is at 1. We know y=4+5(x-1)=5x+3 so the slope of the tangent line is 5. That is f'(1). So, h'(1)=(1/2)(4)^-1/2(5)=5/4.