Find the slope of the normal line to y= ln(15-x) at x=4
To find the slope of the normal line to the curve y = ln(15-x) at x=4, we need to find the derivative of the function at x=4, then find the negative reciprocal of that derivative.
Step 1: Find the derivative of y = ln(15-x).
The derivative of ln(u) with respect to x is 1/u * du/dx. Applying this rule, we can find the derivative of ln(15-x) as follows:
dy/dx = d/dx [ln(15-x)]
= (1/(15-x)) * d/dx (15-x)
Using the power rule, we know that d(x^n)/dx = nx^(n-1). In this case, n = 1, and the derivative of 15-x with respect to x is -1.
dy/dx = (1/(15-x)) * (-1)
= -1/(15-x)
Step 2: Evaluate the derivative at x=4.
Substitute x=4 into the derivative we found in Step 1:
dy/dx = -1/(15-4)
= -1/11
Step 3: Find the negative reciprocal of the derivative.
The negative reciprocal of -1/11 is 11.
Therefore, the slope of the normal line to the curve y = ln(15-x) at x=4 is 11.