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).

we know that f(1) = 4, since the tangent line touches the graph there.

h = √f
dh/dx = 1/(2√f) df/dx

we know that df/dx=5 from the equation of the line. So,

dh/dx = 1/(2√4) * 5 = 5/4

To find h'(1), we need to find the derivative of h(x) and evaluate it at x=1.

First, let's find h'(x), the derivative of h(x). Since h(x) is defined as the square root of f(x), we can use the chain rule to find its derivative.

To do this, we need to find the derivative of the outer function (square root) and the derivative of the inner function (f(x)).

1. Derivative of the outer function:
The derivative of the square root function, √u, is given by:
d/dx (√u) = (1/2√u) * du/dx

2. Derivative of the inner function:
The derivative of f(x), denoted as f'(x), is given in the problem as the equation of the tangent line: f'(x) = 5.

Now, let's find h'(x) using the chain rule:

h'(x) = (1/2√f(x)) * f'(x)

Since we want to find h'(1), we substitute x=1 into the expression for h'(x):

h'(1) = (1/2√f(1)) * f'(1)

Now, let's find f(1) and substitute it into the equation:

Since f(x) is defined as the given equation of the tangent line, we can substitute x=1 to find f(1):

f(1) = 4 + 5(1 - 1) = 4

Now, substitute f(1) and f'(1) into the equation for h'(1):

h'(1) = (1/2√4) * 5

Finally, simplify the expression:

h'(1) = (1/2 * 2) * 5 = 5

Therefore, h'(1) = 5.