1. If f(x) = ln(1 - 2x), then find f '(x)
2. If f(x) = 7x6cos-1x, find f '(x)
1. If f(x) = ln(1 - 2x), then find f '(x)
f' = [ 1/(1-2x) ]( -2)
= -2/(1-2x) = 2/(2x-1)
I do not understand your second function
@damon the 6 and -1 are exponents so
7x^6 and cos^-1
To find the derivative of f(x), we can use the chain rule.
1. If f(x) = ln(1 - 2x), we can find f'(x) as follows:
Step 1: Identify the inner function. In this case, it is 1 - 2x.
Step 2: Find the derivative of the inner function. The derivative of 1 - 2x with respect to x is -2.
Step 3: Apply the chain rule. The derivative of ln(u) with respect to u is 1/u. So, the derivative of ln(1 - 2x) with respect to x is (1 / (1 - 2x)) * -2.
Therefore, f'(x) = -2 / (1 - 2x).
2. If f(x) = 7x^6 * cos^(-1)(x), we can find f'(x) using the product rule and chain rule.
Step 1: Identify the first function, which is 7x^6, and find its derivative. The derivative of 7x^6 with respect to x is 42x^5.
Step 2: Identify the second function, which is cos^(-1)(x). Differentiate cos^(-1)(x) using the chain rule as follows:
a) Let u = cos^(-1)(x).
b) Let y = cos(u).
c) Find dy/du = -sin(u) using the derivative of cos(u).
d) Finally, find dy/dx as dy/du * du/dx = -sin(u) * du/dx.
Wrapping up step 2, the derivative of cos^(-1)(x) with respect to x is -sin(cos^(-1)(x)).
Step 3: Apply the product rule. The product rule states that if f(x) = g(x) * h(x), then f'(x) = g'(x) * h(x) + g(x) * h'(x).
So, f'(x) = (42x^5) * cos^(-1)(x) + 7x^6 * (-sin(cos^(-1)(x))).
Therefore, f'(x) = 42x^5 * cos^(-1)(x) - 7x^6 * sin(cos^(-1)(x)).