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**george****Guest**

Hello all, hope everybody has had a happy new year..

I have a little trouble with this differential question

y=3x^6 + sin2x +e^2x + e^4x log e^2x

i need to find dy/dx.

I would be grateful for an answer

Thanks

**God****Member**- Registered: 2005-08-25
- Posts: 59

y=3x^6 + sin2x +e^2x + e^4x log e^2x

Use power rule for 3x^6 -> 18x^5

Use chain rule for next two terms:

e^2x -> 2e^2x; sin2x -> 2cos2x

so so far that's dy/dx = 18 x^5 + 2 cos2x + 2 e^2x + ...

... can you put parenthases in the last term so that I know what exactly it says?

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**george****Guest**

I knew god would help me....lol

Thanx

If y = 3x^6 + sin2x + e^2x + e^4x log e^2x

Find dy/dx

??

**God****Member**- Registered: 2005-08-25
- Posts: 59

is log natural log or base 10?

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**george****Guest**

it is natural log

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'm going to jump in here and assume that it means e^4x * ln (e^2x).

The e and the ln cancel each other out, so you're left with 2x*e^4x.

And to differentiate that, you use the product rule: (uv)' = uv' + vu'.

d(2x*e^4x)/dx = 8x*e^4x + 2*e^4x = e^4x(8x+2)

Put that together with what God already found, and you get: dy/dx = 18 x^5 + 2 cos2x + 2 e^2x + e^4x(8x+2)

Why did the vector cross the road?

It wanted to be normal.

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