Question

Exp Log Problem

Answer 1

\(\bf f(x)=\cfrac{-2x^6}{ln(x)}\quad ?\)

Answer 2

\[f(x) = -2x^{6} \ln x\]
f'( x ) =
f'( e^ {4} ) =

Answer 3

there

Answer 4

well, for the f'(x) just use the product rule

Answer 5

(-2x^6)(1/x) + (-12x^5)(lnx) = dy/dx

Answer 6

\[-2x^{5}\!\left(6\ln\!\left(x\right)+1\right) \]

Answer 7

(-2x^6)(1/x) + (-12x^5)(lnx) = dy/dx \(\checkmark\)

Answer 8

2nd comment is simplified, how do i get f'(e^4)?

Answer 9

well.... tis simple if you recall the log cancellation rule of \(\bf log_{\color{red}{ a}}{\color{red}{ a}}^x=x\qquad thus \quad ln(e^4)\to log_{\color{red}{ e}}{\color{red}{ e}}^4\iff 4\)

Answer 10

so (-2(e^4)^5)(6*4 + 1)

Answer 11

yeap

Answer 12

e^4)^5 = e^20?

Answer 13

yeap

Answer 14

genius! thank you :)

Answer 15

yw