Question

Find the derivatives of implicit function: e^xy=e^4x-e^5y

Answer 1

\(\large\color{black}{e^{xy}=e^{4x}-e^{5y} }\)

Answer 2

Do want to take the derivatives with e^x, and apply chain rule for the exponents,
or do logarithmic differentiation ?

Answer 3

logarithmic differentiation

Answer 4

\(\large\color{black}{e^{xy}=e^{4x}-e^{5y} }\)
wait actually without logarithmic it is easier...

\(\large\color{black}{e^{xy}\times(x\times \frac{dy}{dx}+y\times 1)=e^{4x}\times4-e^{5y}\times 5\frac{dy}{dx} }\)

Answer 5

I am just applying the chain rule to the exponents.

Answer 6

And then we get,
\(\large\color{black}{e^{xy}\times(x\frac{dy}{dx}+y)=4e^{4x}-5e^{5y} \frac{dy}{dx} }\)

\(\large\color{black}{\frac{dy}{dx}xe^{xy}+ye^{xy}=4e^{4x}-5e^{5y} \frac{dy}{dx} }\)

Answer 7

\(\large\color{black}{\frac{dy}{dx}xe^{xy}+5e^{5y} \frac{dy}{dx}=4e^{4x}-ye^{xy} }\)

Answer 8

\(\large\color{black}{\frac{dy}{dx}(xe^{xy}+5e^{5y})=4e^{4x}-ye^{xy} }\)

Answer 9

\(\LARGE\color{black}{\frac{dy}{dx}=\frac{4e^{4x}-ye^{xy}}{xe^{xy}+5e^{5y}} }\)

Answer 10

you can perhaps simplify but this is the derivative.

Answer 11

Thank you! Very helpful!

Answer 12

I didn't like logarithmic differentiation, because then we would have
\(\LARGE\color{black}{\ln(e^{4x}-e^{5y}) }\) on the right side.

That would mean, chain rule for the natural log, then chain rule for each of the exponents.... to many chains.....

You welcome:)