find the derivative of the function. y = 3^{4^{x^{2}}} y =

Explanation:

Step1: Let $u = 4^{x^{2}}$

$y = 3^{u}$

Step2: Find $\frac{dy}{du}$

Using the formula for the derivative of $a^{u}$ ($a>0,a\neq1$) which is $\frac{d}{du}(a^{u})=a^{u}\ln a$, for $y = 3^{u}$, we have $\frac{dy}{du}=3^{u}\ln 3$.

Step3: Find $\frac{du}{dx}$

Let $v=x^{2}$, then $u = 4^{v}$. First find $\frac{du}{dv}$ and $\frac{dv}{dx}$. For $u = 4^{v}$, $\frac{du}{dv}=4^{v}\ln 4$ by the formula $\frac{d}{dv}(a^{v})=a^{v}\ln a$ with $a = 4$. And $\frac{dv}{dx}=2x$. By the chain - rule $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=4^{x^{2}}\ln 4\cdot2x$.

Step4: Find $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=3^{u}\ln 3$ and $\frac{du}{dx}=4^{x^{2}}\ln 4\cdot2x$ into it. Since $u = 4^{x^{2}}$, we get $\frac{dy}{dx}=3^{4^{x^{2}}}\ln 3\cdot4^{x^{2}}\ln 4\cdot2x$.

Answer:

$2x\cdot4^{x^{2}}\ln 4\cdot3^{4^{x^{2}}}\ln 3$