Question

find dy/dx for y = ln (7x^3 - x^2)

Answer 1

(lny)'=(1/y)*(dy/dx)

Answer 2

after you get 1 / 7x^3-x^2, how do you solve that?

Answer 3

u multiply that by (7x^3-x^2)'

Answer 4

But keep in mind that before differentiating, we must remember that the domain for this function is \(\bf x > \frac{1}{7}\).

Then do what @bahrom7893 did. You are basically applying the chain rule here. After you get 1/7x^3 - x^2, multiply it by the derivative of whatever is inside the bracket.

Answer 5

what do you mean?

Answer 6

the derivative of that?
so 21x^2 - 2x ?

Answer 7

yes that is correct.

Answer 8

then what do you do?

Answer 9

multiply the 21x^2 - 2x by the 1/7x^3 - x^2

Answer 10

so how do you do that....?

Answer 11

Remember, this is the chain rule. We take the derivative as 1/7x^3-x^2. But because 7x^3 - x^2 is itself a function, we must now multiply by its derivative hence:\[\bf \frac{d}{dx}\ln(7x^3-x^2)=\frac{ 1 }{ 7x^3-x^2} \times \frac{ d }{ dx }7x^3-x^2\]\[\bf = \frac{ 1 }{ 7x^3-x^2 } \times 21x^2-2x=\frac{ 21x^2-2x }{ 7x^3-x^2 }\]

Answer 12

You can factor out the x from the top and bottom and simplify it further but that's the idea. \[\bf \frac{ d }{ dx } \ln(g(x))=\frac{ 1 }{ g(x) } \times g'(x)\]And that's exactly what we did. @gbrooke21 Get it?

Answer 13

I see, for the bottom can you simplify the denominator?

Answer 14

Can it be
21x-2/7x^2 - x?

Answer 15

@gbrooke21 Yes that's right. Factor the x out from the top and bottom and that's what you will get. Good job.