how to differentiate this equation: Y = 3^(2x^4)?

Question

thanks.

Answer 1

: Y = 3^(2x^4)
lnY = ln(3^(2x^4))
= 2x^4 ln(3)
so Y = e^{2x^4 ln(3)}

dY/dx = e^{2x^4 ln(3)} d/dx {2x^4 ln(3)}
= e^{2x^4 ln(3)} 8x^3 ln(3)
= 3^(2x^4) (8x^3) ln(3)

Answer 2

take the natural log of both sides to get
ln(y)=(2x^4)*ln(3)
(take the derivative)
y'/y=8x^3*ln(3)
y' (or dy/dx)=y(8x^3*ln(3))
and using the y value from the initial problem,
y'=8x^3*ln(3)*3^(2x^4)
or at least I think that's right... this stuff is right about where I'm at in math

Answer 3

y = a^x => dy/dx = (a^x)*ln(a) so
dy/dx = (3^(2x^4))*ln(3)* 8x^3
and you should be able to take it from there ☺

HTH

Doug