Question

Derivative. variable in base and exponent.
f(x)=x^{2x+3}
I haven't been able to find the steps to solve a derivative with x in the base and the exponent. How do I solve this type of derivative?

Answer 1

\[f(x)=x ^{2x+3}\]
(formatted)

Answer 2

best way is to use logs
\[y = x^{2x+3}\]
\[\ln y = (2x+3) \ln x\]
now differentiate
\[\frac{1}{y} \frac{dy}{dx} = \frac{2x+3}{x} +2\ln x\]
\[\frac{dy}{dx} = x^{2x+3}(\frac{2x+3}{x} +2\ln x)\]

Answer 3

so rewrite it as "y=", take ln, then use implicit differentiation?

Answer 4

correct

Answer 5

I see. dy/dx trips me up sometimes, so I stick to y' and looks like I get the same answer.
\[y=x^{2x+3}\]
\[\ln (y) = (2x+3) \ln(x)\]

then take derivative of both sides.
\[\frac{ y' }{ y }=2\ln(x)+(2x+3)\frac{ 1 }{ x }\]
\[y'=y(2\ln(x)+\frac{ 2x+3 }{ x })\]
\[y'=x^{2x+3}(2\ln(x)+\frac{ 2x+3 }{ x })\]
Thanks!

Answer 6

yw