Differentiate how? PART 1

Calculus Level 3

If ${ x }^{ y }={ e }^{ x-y }$ , then $\frac { dy }{ dx } = \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}.$

\frac { x-y }{ (1+\log { x)^{ 2 } } }

\frac { \log { x } }{ (1+\log { x)^{ 2 } } }

\frac { x-y }{ 1+\log { x } }

\frac { y-x }{ x(1+\log { x) } }

2 solutions

Niranjan Khanderia
Jun 1, 2015

$\large x^y= e^{ x-y}=\dfrac{e^x}{e^y}~\implies~(ex)^y=e^x,~~\therefore~y*ln(ex)=x,~\\\implies~y= \dfrac x { ln(ex) }=\dfrac x { 1+ lnx} \\ \dfrac { dy }{ dx } =\dfrac 1 { 1+ lnx}- \dfrac x { (1+ lnx)^2}*(0+\frac 1 x ) ............Chain ~Rule.\\ \dfrac { dy }{ dx } = \dfrac{lnx} { (1+ lnx)^2}$

I agree with this method and result, and I wonder why the "correct answer" has it in terms of log x and not ln x?

Wooil Jung - 6 years ago

log x can be assumed as a log with base 'e'

andrews ganj - 6 years ago

I was under the assumption that log x is generally assumed to be log of x to the base of 10, maybe this is only applied to high school levels?

Wooil Jung - 5 years, 11 months ago

@Wooil Jung – log x is generally assumed to be base 10, ln x is base e, and lg x is base 2

Richard Landry - 4 years, 3 months ago

Shreyansh Singh
Jul 7, 2015

Differentiate how? PART 1

2 solutions

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