Question

Hello

Can someone correct this math problem for me? (the file is attached in second post).

Thanks in advance.

Answer 1

Sorry can't help you there

Answer 2

Seems fine to me.

Answer 3

Actually, a teacher told me that my answer wasn't totally correct. But he didn't say what I did wrong. Would you please double-check it?

Answer 4

Ok, sure...

Answer 5

Would you mind if I type everything here?

Answer 6

Of course!

Answer 7

Of course not, I mean haha

Answer 8

\(\rm \Large\color{black}{ \displaystyle m=\ln(qh-2h^2)+2e^{(q-h^2+3)^4}-7 }\)

Answer 9

\( \large\color{black}{ \displaystyle m=\ln(q\color{red}{h}-2\color{red}{h}^2)+2e^{(q-\color{red}{h}^2+3)^4}-7 }\)

\( \large\color{black}{ \displaystyle \partial m/\partial \color{red}{h}=\frac{q-4\color{red}{h}}{q\color{red}{h}-2\color{red}{h}^2} +2e^{(q-\color{red}{h}^2+3)^4}\times4(q-\color{red}{h}^2+3)^3\times (-2h) }\)

Answer 10

(you said +2h)

Answer 11

Yes, thank you :) But what about the the dm/dq part?

Answer 12

Let me take a look at that....

Answer 13

\(\large\color{black}{ \displaystyle m=\ln(\color{blue}{q}h-2h^2)+2e^{(\color{blue}{q}-h^2+3)^4}-7 }\)

\(\large\color{black}{ \displaystyle \partial m/\partial \color{blue}{q}=\frac{h}{\color{blue}{q}h-2h^2} +2e^{(\color{blue}{q}-h^2+3)^4}\times 4{(\color{blue}{q}-h^2+3)^3}\times 1 }\)

Answer 14

this ×1, is the chain rule.
(To make sure it is understood that there is really no chain rule 2nd time by dm/dq)

Answer 15

Now it makes sense :) Thank you so much once again!

Answer 16

New variables, m h and q :)

You welcome...