Question

Find the derivative of 3^(-x^2)

Answer 1

\[\frac{ d}{ dx} \left[ 3-^{x ^{2}}\right]\]

Answer 2

I thought the derivative of a^u is (a^u)(ln a)(du/dx) so I got
3^(-x^2) (ln3) (-2x) but the answer is -3^(-x^2)xln9

Answer 3

http://www.derivative-calculator.net/
calculator for it ^

Answer 4

Their method is too long and involves setting all of the input derivatives equal to imaginary y value and then making the problem in terms of e^logx whatever. Imo it's useless.

Answer 5

Or needlessly complex and difficult to understand, rather.

Answer 6

Let y = 3^(-x^2)

Apply the natural log to both sides


ln(y) = ln[ 3^(-x^2) ]
ln(y) = -x^2*ln(3)


Now derive both sides with respect to x and isolate dy/dx

Answer 7

Is there a way to do it without implicit differentiation?

Answer 8

you can use ` derivative of a^u is (a^u)(ln a)(du/dx) `

u = -x^2
du/dx = -2x
a = 3

y = a^u
dy/dx = (a^u)(ln a)(du/dx)

so if
y = 3^(-x^2)
then
dy/dx = (a^u)(ln a)(du/dx)
dy/dx = 3^(-x^2)*ln(3)*(-2x)
dy/dx = -3^(-x^2)*x*2ln(3)
dy/dx = -3^(-x^2)*x*ln(3^2)
dy/dx = -3^(-x^2)*x*ln(9)

Answer 9

Oh. I forgot that you could apply the log power rule here. Thank you!

Answer 10

you're welcome

Answer 11

Here is how to do it using implicit differentiation


Let y = 3^(-x^2)
ln(y) = ln[ 3^(-x^2) ]
ln(y) = -x^2*ln(3)
y'/y = -2x*ln(3)
y' = y*(-2x*ln(3))
y' = -3^(-x^2)*2x*ln(3)
y' = -3^(-x^2)*x*2ln(3)
y' = -3^(-x^2)*x*ln(3^2)
y' = -3^(-x^2)*x*ln(9)


so we get the same answer