# Logarithmic differentiation

f(x)=(2x-1)^-2 (x+3)^4 at x=1

f'(x)= -2*2/2x+1 + 4*1/x+3 [(2x-1)^-2)(x+3)^4 and then I plug in 1.

I am having trouble getting an answer when I plug in 1.

## Can you re-write I think there is something missing.

## You lost me totally

d/dx (a * b) = a db/dx + b da/dx

f= (2x-1)^-2 * (x+3)^4

f' = [(2x-1)^-2][4(x+3)^3]

+ (x+3)^4 (-2(2x-1)^-3 (2)]

at x = 1

f' = [(2-1)^-2][4(4)^3]

+ (4)^4 (-2(1)^-3 (2)]

which is

f' = [1][4^4]

+ (4)^4 (1/4]

or

4^4+4^3 = 4^3(5)= 320

check my arithmetic !

## To find the derivative of the function f(x) = (2x-1)^-2 (x+3)^4 at x = 1 using logarithmic differentiation, follow these steps:

1. Take the natural logarithm (ln) of both sides of the equation f(x) = (2x-1)^-2 (x+3)^4. This step allows us to use logarithmic properties, making the differentiation easier. So, we have: ln(f(x)) = ln((2x-1)^-2 (x+3)^4).

2. Apply the logarithmic rule for multiplication: ln(a * b) = ln(a) + ln(b). Using this rule, we can rewrite the equation as ln(f(x)) = ln((2x-1)^-2) + ln((x+3)^4).

3. Use the exponent rule for logarithms: ln(a^n) = n * ln(a). Applying this rule to our equation, we have ln(f(x)) = -2 * ln(2x-1) + 4 * ln(x+3).

4. Differentiate both sides of the equation with respect to x. The left side will be the derivative of ln(f(x)) with respect to x, which is (1/f(x)) * f'(x). The right side will be the derivative of the expression on the right-hand side of the equation.

5. Apply the chain rule for differentiation when finding the derivative on the right-hand side. The derivative of ln(g(x)) with respect to x is g'(x)/g(x). So, the derivative of -2 * ln(2x-1) with respect to x is -2 * [(d/dx) (2x-1)] / (2x-1). Similarly, the derivative of 4 * ln(x+3) with respect to x is 4 * [(d/dx) (x+3)] / (x+3).

6. Simplify the derivatives on the right side. The derivative of (d/dx) (2x-1) is simply 2, and the derivative of (d/dx) (x+3) is 1.

7. Substitute x = 1 into both the function and its derivative expression. This step will give you the value of f(1) (which is needed for the left side) and f'(1) (which is needed for the right side).

So, let's apply these steps to our problem:

Step 1: ln(f(x)) = ln((2x-1)^-2 (x+3)^4).

Step 2: ln(f(x)) = ln((2x-1)^-2) + ln((x+3)^4).

Step 3: ln(f(x)) = -2 * ln(2x-1) + 4 * ln(x+3).

Step 4: (1/f(x)) * f'(x) = -2 * (2/(2x-1)) + 4 * (1/(x+3)).

Step 5: (1/f(x)) * f'(x) = -4/(2x-1) + 4/(x+3).

Step 6: Now, plugging in x = 1 into the expression on the right side:

-4/(2(1)-1) + 4/(1+3) = -4 + 1 = -3.

So, the derivative of f(x) = (2x-1)^-2 (x+3)^4 at x = 1 is -3.