Feb 5, 2017 · d/dx (e^(ex)lnx) = e^(ex)(elnx +1/x) Using the product rule: (1) d/dx (e^(ex)lnx) = d/dx ( e^(ex))lnx +e^(ex)d/dx(lnx) Now we have using the chain rule: d/dx ( e^(ex ...

May 8, 2016 · #color(red)(ln(x))# means #color(red)(" the value "k" you need as an exponent of e for " e^k" to be equal to x")#

Apr 3, 2018 · lne^lnx=lnx As lna^b=b*lna, lne^lnx=lnx*lne but as base in lnx is e, lne=1 hence lne^lnx=lnx

Dec 21, 2016 · y' = 2x "e" and the natural logarithms are inverses, so they equal x when in the form ln(e^x) or e^(lnx). Thus, y = x^2 The derivative of this is obtained using the power rule. y'= 2x Hopefully this helps!

Jan 25, 2017 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . Andrea S.

Jul 28, 2017 · If f(x) =-e^(-3x-7) and g(x) = lnx^2 , what is f'(g(x)) ? Given f(x) =-e^(-3x-7) Differentiating w . r to x we gat f'(x) =-e^(-3x-7)xx(d/(dx) (-3x-7)) =>f'(x) =3e ...

Jun 17, 2016 · x^3 You can write: e^(3lnx)=(e^lnx)^3 =x^3

Feb 10, 2016 · Use e^lnx = x. Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Calculators

Apr 23, 2018 · Use your log laws... #ln alpha ^ beta = beta ln alpha # Take #ln# both sides... #=> lnx^(lnx) = ln3^9 # #=> lnx * lnx = 9ln3 #

Aug 16, 2016 · I would say FALSE. Consider it: e^(6lnx)= let us focus our attention on the exponent. We can use the property of logs to write it as: =e^(lnx^6)= now we use the definition of log and the fact that e and ln eliminate each other to give: x^6, or: =cancel(e)^(cancel(ln)x^6)=x^6