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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
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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I knew god would help me....lol
Thanx
If y = 3x^6 + sin2x + e^2x + e^4x log e^2x
Find dy/dx
??
is log natural log or base 10?
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it is natural log
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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