Find the derivative of f(x) = 3^(x^2+1)?

Question

I'm confused bc its not zero, right?

Answer 1

If f(x) = 3^(x² + 1), then, by taking natural logs of both sides:

ln{ f(x) } = ln{ 3^(x² + 1) } = (x² + 1) ln{ 3 }

Differentiate both sides with respect to x:

1/ f(x) f ' (x) = (2x) ln{ 3 }

⇒ f ' (x) = f(x) (2x) ln{ 3 } and, on replacing the f(x) by its original expression from line 1 above,

f ' (x) = 2.ln{ 3 }.x. 3^(x² + 1)

Hope it's now clear.

Answer 2

I think mitch l is right.

d(a^x)/dx = a^x ln a, but x here is a function of x, so chain rule:

d[ 3^(x² + 1)]/dx = 3^(x²+1) • ln 3 • d(x²+1)/dx = 3^(x²+1) • ln 3 • 2x

Answer 3

f(x) = 3^(x^2+1)=3*3^(2x)

derivative = 3*3^(2x)*ln3

Answer 4

ln(3)*2x *3^(x^2+1)

i think im not sure :P

Answer 5

Let y=f(x),
then lny=ln(3^(x^2+1)).
so lny=(x^2+1)ln3.
d(lny)/dx=d[(x^2+1)ln3]/dx,
y^(-1)dy/dx=2xln3
dy/dx=2x(ln3)*y=2x(ln3)*3^(x^2+1).