Given that y = e^f(x), show that y' = f'(x)e^f(x).
Hence, find y' when f(x) = x.
Given that y = e^f(x), show that y' = f'(x)e^f(x).
Hence, find y' when f(x) = x.
this is just the chain rule. I assume you have seen the proof of that.
That said, just substitute x for f(x).
sorry sir pleasee show me how
There are lots of proofs of the chain rule on line. You can find one here, if you scroll down some
http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx
So, now you know that if
y = f(g) and g is a function of x, then
dy/dx = df/dg * dg/dx
You can see that treated as fractions, the "dg" factors cancel.
Now, how do you know that
d/dx = e^x ?
There are lots of proofs of that, as well, such as
http://math.stackexchange.com/questions/190773/proof-of-fracddxex-ex
(or, probably, in your own math text!)
So, if y = e^f(x)
y' = e^f f'(x)
If f = x, then f'=1, and we have the very useful result that
d/dx e^x = e^x
To prove the given statement, we'll use the chain rule for differentiation.
Let's start by differentiating both sides of the equation y = e^f(x) with respect to x:
dy/dx = d/dx(e^f(x))
To differentiate the right side of the equation, we'll use the chain rule.
Let u = f(x) and v = e^u. Thus, y = v.
Applying the chain rule, we have:
dy/dx = dv/du * du/dx
To find dv/du, we differentiate v = e^u with respect to u.
dv/du = d/du(e^u) = e^u
Now, let's differentiate u = f(x) with respect to x to find du/dx.
du/dx = d/dx(f(x)) = f'(x)
Substituting the values back into the chain rule expression, we get:
dy/dx = e^u * f'(x)
Since y = v, we can substitute back in e^f(x) for v:
dy/dx = e^f(x) * f'(x)
Therefore, we have shown that y' = f'(x) * e^f(x).
To find y' when f(x) = x, we substitute f(x) = x into the derived result:
y' = f'(x) * e^f(x)
y' = 1 * e^x
y' = e^x
Hence, when f(x) = x, y' = e^x.