Question

Use logarithmic differentiation to find dy/dx if
y = (x^3 - )^5 / e^7x sin (2x).


the given answer = (x^3-1)^5/e^7x sin (2x) [ 15x^2 /x^3-1 -7 - 2 cot 2(2x)]

Answer 1

you have to use quotient rule

Answer 2

I deserve a medal!!!!!!!!!!!!!11

Answer 3

i have try to do it but how to remove e^7x

Answer 4

No, don't use the quotient rule since you were asked to use logarithmic differeniation.

Answer 5

forget whatever i said

Answer 6

it okay KissMyAxe

Answer 7

Logarithm differentiation means to take the logarithm of both sides of the equation and then take the derivative. Here's an example: \[\Large y=\frac{(x+7)^6}{ e^{2x}} \\ \Large \ln y = 6 \ln (x+7) -2x \] See, through using log rules you can split it apart, then take the derivative. Then you'll have
\[\Large \frac{y'}{y}=6\frac{1}{x+7}-2\] But we want y' so we multiply y to both sides and then plug it in from earlier! Done!
It's ok @KissMyAxe =)

Answer 8

I get it @Kainui

Answer 9

In y = In (x^3-1)^5 e^(-7x) / In (sin 2x)^1
ln y = 5 ln (x^3-1) - 7x - 1 ln ( sin 2x)

(1/y) dy/dx = 5 (3x^2) /(x^3-1) - 7 - (1 /sin 2x) cos (2x) (2)
(1/y) dy/dx = 15x^2 /(x^3-1) - 7 - 2 (cos 2x / sin 2x)
(1/y) dy/dx = 15x^2 /(x^3-1) - 7 - 2 cot(2x)

dy/dx = y [15x^2 /(x^3-1) - 7 - 2 cot(2x) ]
dy/dx = (x^3-1)^5 / e^7x sin(2x) ( 15x^2 /(x^3-1) - 7 - 2 cot(2x) )

tq for helping me