Question

Find dy/dx if f(x) = (x+1)^(2x)

Answer 1

Logarithmic differentiation

Answer 2

\(\color{#000000 }{ \displaystyle y=(x+1)^{2x} }\)

hit both sides with a natural log, and then differentiate on both sides.

Answer 3

Or,

Answer 4

\(\color{#000000 }{ \displaystyle y=(x+1)^{2x} =e^{\ln(x+1)^{2x}}=e^{2x\ln(x+1)}}\)

Answer 5

Here we can make use of the derivation rule:
\[f(x)=A^n \iff f'(x)=(n)A ^{n-1}\]
Now, this will be okay, but thing is, what we have in front of us is what we call a composite function, since the racial is also a function, we might want to conveniently transport it to the numerator by applying a logarithm.
Or, as I would do it:
\[f(x) = (x+1)^{2x} \iff f(x)=e ^{\ln(x+1)^{2x}}\]
Which gets traduced to:
\[f(x)=e ^{(2x)\ln(x+1) }\]
And that is most easy to derive, after all, the drivative of:
\[h(x)=e ^{f(x)} \iff h'(x)= e ^{f(x)}.f'(x)\]

Answer 6

let \(\color{#000000 }{ \displaystyle f }\) and \(\color{#000000 }{ \displaystyle g }\) be differentiable functions of x, and let \(\color{#000000 }{ \displaystyle z=f^g }\).

We can do a little algebra;
\(\color{#000000 }{ \displaystyle z=f^g}\)
\(\color{#000000 }{ \displaystyle z=e^{\large \ln(f^g)} }\)
\(\color{#000000 }{ \displaystyle z=e^{\large ~g\ln(f)} }\)

Then the derivative
\(\color{#000000 }{ \displaystyle \frac{dz}{dx}=e^{\large ~g\ln(f)}\times \left[ g'\ln(f)+g\frac{f'}{f}\right] }\)

Answer 7

In your case, as you will see (when you come online).

f = (x+1)
g = 2x

Answer 8

saying over the same thing the third time in a different form.

Answer 9

Or use the damn formula for function power function:

y = n^y

y' = y n^(y-1) * n' + n^y ln(n) * y'

the same as taking ln, differentiating and then multiplying times the y ( as Solomon said).