How do you find the derivative of #y=e^(e^(3x^2))#?

Question

Scarlett Allison · Accepted Answer

#y'=6xe^(e^(3x^2)+3x^2)#

The rule for differentiating e functions is,

#y=e^f(x)#

#y'=f'(x)e^f(x)#

so in the case of your question we have

#y=e^f(x)# where #f(x) = e^g(x)#

so we will differentiate to find each part we need, #y=e^f(x)#

#f(x) = e^g(x)#

#f'(x)=g'(x)e^g(x)#

So now what is y, f(x) and g(x).

#y=e^(e^(3x^2))#

#f(x)=e^(3x^2)#

#g(x)=e^(3x^2)#

#f(x)=e^(3x^2)#

#f'(x)= 6x*e^(3x^2)#

so subbing into the original formula,

#y'=f'(x)e^f(x)#

#y'=(6x*e^(3x^2))(e^(e^(3x^2))) #

#y'=6x*e^(e^(3x^2))*e^(3x^2)#

#y'=6xe^(e^(3x^2)+3x^2)#

William Abdalla · Answer

undefined To find the derivative of $ y = e^{e^{3x^2}} $, we use the chain rule. The chain rule states that if we have a function within another function, we take the derivative of the outer function with the inner function intact, then multiply by the derivative of the inner function.

Let $ u = e^{3x^2} $. Then, $ y = e^u $.

Taking the derivative of $ y $ with respect to $ x $ using the chain rule:

$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$

Where:

$$ \frac{dy}{du} = e^u $$
$$ \frac{du}{dx} = \frac{d}{dx}(e^{3x^2}) = 6x \cdot e^{3x^2} $$

Therefore:

$$ \frac{dy}{dx} = e^{e^{3x^2}} \cdot (6x \cdot e^{3x^2}) = 6x \cdot e^{3x^2} \cdot e^{e^{3x^2}} $$

So, the derivative of $ y $ with respect to $ x $ is $ 6x \cdot e^{3x^2} \cdot e^{e^{3x^2}} $.