If e^f(x) = 1 + x^2, then f'(x) =
e^y = 1+x^2
e^y y' = 2x
y' = 2x/e^y = 2x/(1+x^2)
or, since e^y = 1+x^2
y = ln(1+x^2)
y' = 2x/(1+x^2)
Oh, so you're trying to bring a little bit of math into my world, huh? Well, let's see what we can do with this fun little equation!
If we have e^f(x) = 1 + x^2, we can start by taking the natural logarithm of both sides to get:
ln(e^f(x)) = ln(1 + x^2)
Now, since ln(e^f(x)) is just f(x), our equation simplifies to:
f(x) = ln(1 + x^2)
Finally, to find f'(x), we can differentiate this expression with respect to x using the chain rule. But, you know what? I'm feeling a little mischievous today, so instead of giving you the whole step-by-step process, I'll just give you the answer right away:
f'(x) = (2x) / (1 + x^2)
Oh, maths... always trying to make things more serious! But hey, don't worry, I'm here to clown around and make it a little less intimidating.
To find the derivative of f(x), we can use logarithmic differentiation.
1. Take the natural logarithm (ln) of both sides of the equation e^f(x) = 1 + x^2:
ln(e^f(x)) = ln(1 + x^2)
2. Apply the logarithmic property that ln(e^a) = a:
f(x) = ln(1 + x^2)
3. Now, we can differentiate both sides of the equation with respect to x. The derivative of ln(1 + x^2) can be found using the chain rule:
f'(x) = (1/(1 + x^2))(2x)
Simplifying further:
f'(x) = (2x)/(1 + x^2)
So, f'(x) = (2x)/(1 + x^2) is the derivative of f(x) with respect to x.
To find the derivative of f(x), we can use the chain rule and the fact that the derivative of e^u is u' * e^u.
Given that e^f(x) = 1 + x^2, we can rewrite it as e^f(x) - 1 = x^2.
Taking the natural logarithm of both sides, we have ln(e^f(x) - 1) = ln(x^2).
Using the properties of logarithms, we get f(x) - 1 = 2ln(x).
Now, let's solve for f(x):
f(x) = 2ln(x) + 1.
To find f'(x), we can differentiate the equation with respect to x:
f'(x) = 2 * (1/x).
Therefore, f'(x) = 2/x.