If f(x)=(x^2+1)^(2-3x), then f'(1) = -3/2ln(2)-1/2
Is this correct?
I'd say no. It's just an exponential function.
No way any logs can creep in there.
So -1/2
To find the derivative of the function f(x) = (x^2+1)^(2-3x), you'll need to use the chain rule. The chain rule states that if you have a composite function y = f(g(x)), then the derivative dy/dx is given by dy/dx = f'(g(x)) * g'(x).
First, let's find the derivative of the outer function f(g(x)) = g^(2-3x). To do this, we can use logarithmic differentiation. Taking the natural logarithm of both sides, we have ln(f(x)) = (2-3x) * ln(g(x)). Now, we can differentiate both sides with respect to x using the product rule:
d[ln(f(x))]/dx = [(2-3x) * ln(g(x))]'
Using the product rule, the right side can be expanded as:
d[ln(f(x))]/dx = (2-3x)'*ln(g(x)) + (2-3x)*(ln(g(x)))'
Since ln(g) is a function of x, we need to differentiate it using the chain rule. Let's define u = g(x). Then ln(g(x)) = ln(u). The derivative of ln(u) with respect to u is 1/u, and multiplying by the derivative of u with respect to x gives:
(ln(g(x)))' = (1/u) * (du/dx)
= g'(x) / g(x)
Substituting this result back into the equation, we have:
d[ln(f(x))]/dx = (2-3x)'*ln(g(x)) + (2-3x)*(g'(x) / g(x)).
Now, let's find the derivative of the inner function g(x) = x^2 + 1. Taking the derivative of g(x) with respect to x gives:
g'(x) = 2x.
Substituting this back into the previous equation, we have:
d[ln(f(x))]/dx = (2-3x)'*ln(g(x)) + (2-3x)*(2x / g(x)).
To find the value of f'(1), we need to evaluate this expression at x = 1. Let's substitute x = 1 into the equation and simplify:
d[ln(f(x))]/dx = (2-3*1)'*ln(g(1)) + (2-3*1)*(2*1 / g(1))
= (-1)*ln(g(1)) + (2-3)*(2 / g(1))
= -ln(g(1)) - 2 / g(1).
Now, let's find g(1) by substituting x = 1 into g(x):
g(1) = 1^2 + 1
= 2.
Substituting this back into the previous equation, we have:
d[ln(f(x))]/dx = -ln(2) - 2/2
= -ln(2) - 1.
Therefore, f'(1) = -ln(2) - 1, not -3/2ln(2)-1/2.