Differentiate the function f(x) = In(2x +3)
d (ln z)/dx= 1/z * dz/dx
Try that, I will be happy to critique your work.
25+(-15)
To differentiate the function f(x) = ln(2x + 3), we can use the rules of differentiation for the natural logarithm function.
The basic rule for differentiating ln(x) is as follows:
d/dx[ln(x)] = 1/x
However, in this case, we have ln(2x + 3), which is not in the form of just ln(x). To handle this, we can use the chain rule of differentiation. The chain rule states that if we have a function of the form g(f(x)), then the derivative is given by:
d/dx[g(f(x))] = g'(f(x)) * f'(x)
Applying the chain rule to our function, we have:
f(x) = ln(2x + 3)
g(u) = ln(u) (where u = 2x + 3)
To find g'(u), we differentiate ln(u) with respect to u:
g'(u) = 1/u
Now, we need to find f'(x), which is the derivative of (2x + 3) with respect to x:
f'(x) = 2
Finally, we substitute the derivatives back into the chain rule formula:
d/dx[ln(2x + 3)] = g'(u) * f'(x)
= (1/u) * 2
= 2/(2x + 3)
Therefore, the derivative of f(x) = ln(2x + 3) is:
f'(x) = 2/(2x + 3)