Question

Using logarithmic differentiation, differentiate: z=(((2x-3)^5)((4x+5)^(8)))/(((x+5)^(7))((3x^(4)-2x^(2)+7)^(9))(e^(4x))

Answer 1

The number of brackets are too sick, im turning blind.

Answer 2

\[\large \mathsf{z=\frac{(2x-3)^5(4x+5)^8}{((x+5)^7(3x^4-2x^2+7)^9(e^{4x})}}\]

Answer 3

yes that's right

Answer 4

That's gonna be long

Answer 5

start by ln everything from the top and bottom of the right side

Answer 6

from there you can use the quotient rule

Answer 7

\[\large \mathsf{\ln z = 5\ln(2x-3) + 8\ln (4x+5) -7\ln(x+5) -9ln(3x^4-2x^2+7) - 4x }\]

\[\mathsf{\frac{dz}{dx} = z \frac{d\left(5\ln(2x-3) + 8\ln (4x+5) -7\ln(x+5) -9ln(3x^4-2x^2+7) - 4x \right)}{dx}}\]

I won't help further, I have done enough work for a medal.

Answer 8

\[ \mathsf{\ln z = 5\ln(2x-3) + 8\ln (4x+5) -7\ln(x+5) -9ln(3x^4-2x^2+7) - 4x }\]

Answer 9

don't you have to multiply that whole thing by the original?

Answer 10

Yeah, that is what 'Z' in RHS represents. If you I can do the whole originial thing too but I don't think you will be able to see the equation then.

Answer 11

If you want* I ...

Answer 12

Yes please

Answer 13

\[\mathsf{\frac{dz}{dx} = \frac{(2x-3)^5(4x+5)^8}{(x+5)^7(3x^4-2x^2+7)^9(e^{4x})} \frac{d\left(5\ln(2x-3) + 8\ln (4x+5) -7\ln(x+5) -9ln(3x^4-2x^2+7) - 4x \right)}{dx}}\]


\[\small \mathsf{\frac{dz}{dx} = \frac{(2x-3)^5(4x+5)^8}{(x+5)^7(3x^4-2x^2+7)^9(e^{4x})} \frac{d\left(5\ln(2x-3) + 8\ln (4x+5) -7\ln(x+5) -9ln(3x^4-2x^2+7) - 4x \right)}{dx}}\]