Differentiate y = 3sin^2 x e^5x – 1 with respect to x

Original post by zeesus699

Can anyone differentiate y = 3sin^2 x e^5x – 1 with respect to x? I have completed the question, however as i don't have the answer to refer to i am not sure its 100% correct.

Any help?

Amended due to ambiguous notation:

http://www.wolframalpha.com/input/?i=differentiate+y+%3D+3sin%5E2x+e%5E(5x-1)

(edited 7 years ago)

Reply 2

http://www.wolframalpha.com/input/?i=differentiate+y+%3D+3sin%5E2x+e%5E(5x)-1

Original post by Mr M

Absolute legend!

Reply 3

Original post by zeesus699

Absolute legend!

Remember the website. It is great for checking your work.

Reply 4

It turns out my working out are wrong, and i can't seem to find the correct method of producing that answer. Any guidance?

Reply 5

Original post by zeesus699

It turns out my working out are wrong, and i can't seem to find the correct method of producing that answer. Any guidance?

Post your working and someone will tell you where the error is.

Reply 6

So far, using the product rules.

y = 3sin^2 x e^5x – 1
U=e^5x-1
V=3sin^2x

du/dv= e^5x-1
dv/du=6sin(x)cos(x)

Dy/dx= u*(dv/du) + V*(du/dx)

Dy/dx= e^5x-1 6sin(x)cos(x) + 3sin^2x e^5x-1

Any help in where i am going wrong? I am relatively new to doing this level of differentiating.

Reply 7

Original post by zeesus699

Ok I can see the problem. Your notation was ambiguous. I interpreted your question as

e^{(5x)}-1

rather than

e^{(5x-1)}

. I'll amend the link to Wolfram. Also your

\frac{du}{dx}

(NOT

\frac{du}{dv}

) is wrong.

(edited 7 years ago)

Reply 8

Original post by Mr M

Ok I can see the problem. Your notation was ambiguous. I interpreted your question as

e^{(5x)}-1

rather than

e^{(5x-1)}

. I'll amend the link to Wolfram. Also your

\frac{du}{dx}

(NOT

\frac{du}{dv}

) is wrong.

Much appreciated, however I have already amended it beforehand. Oh, and the "du/dv" was a typing error. Thanks for the correction.

I am having trouble trying to figure out how to differentiate more complex trigonometric functions more than anything, as my knowledge of them is not 100%. How would I go about trying to differentiate this?

Thanks

Reply 9

Original post by zeesus699

The trigonometric bit was correct. The exponential bit was incorrectly differentiated.

Unparseable latex formula:

\diplaystyle \frac{d}{dx} e^{ax+b} = a e^{ax+b}

Reply 10

Gurtajs

Excuse my bad hand writing

Original post by Gurtajs

...

Please read the Forum Rules at the top of the Maths Forum. Full worked solutions are not allowed. Just give hints please.

Reply 13

Original post by Olmeister

...

Please read the Forum Rules at the top of the Maths Forum. Full worked solutions are not allowed. Just give hints please.

Reply 14

Olmeister

Original post by Mr M

Please read the Forum Rules at the top of the Maths Forum. Full worked solutions are not allowed. Just give hints please.

I have never seen these rules, and apologise as I did not realise that this is what the standards are. However, I feel that personally when I do post a question on here it is easier to get someone else's working and then compare it to my own. It gives me practice as well as the person asking the question.

I will keep to only posting hints in the future.

Reply 15