Calculus problem, Use logarithmic differentiation to find dy/dx for y = (3x-4)/(2x+5)^4

Hey all, need some help checking my solution for this problem:

y = (3x-4)/(2x+5)^4

I'll go into more detail..

y = (3x-4)/(2x+5)^4

when using logs, division switches to subtraction

ln y = ln (3x-4) - ln (2x+5)^4

Also when using logs the exponent can go in front of the log and be multiplied by it.

ln y = ln (3x-4) - 4*ln (2x+5)

Now when differentiating a natural log, use following formula. lnx ----> 1/x * derivative of x.

y'/y = 1 / (3x-4) *3 - 4*(1/(2x+5))*2

Just make it look better....

y'/y = 3/(3x-4) - 8/(2x+5)

Multiply y over and resubstitute the y back in (original equation).

y' = (3x-4)/(2x+5)^4 [ 3/(3x-4) - 8/(2x+5)]

I would really suggest leaving it at that.

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It's not
4*(1/(2x+5)^2

it is

4*(1/(2x+5)) * 2

Notice the multiplication, it is not an exponent. So it is actually 4*2*1 which is 8. Hope it helps a little.

y = (3x-4)/(2x+5)^4
ln y = ln(3x - 4) - 4ln(2x + 5)
y'/y = 3/(3x - 4) - 8/(2x + 5)
y' = (3x-4)/(2x+5)^4 [3/(3x - 4) - 8/(2x + 5)]

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ln(x)^y = y ln(x)

Hence ln(2x + 5)^4 = 4ln(2x + 5)
Use chain rule derivative of 4ln(2x + 5) = 4/(2x + 5) * 2 = 8/(2x + 5). That is where the 8 comes from.