find the derivative of the function\n\n$y = e^{(x^{2}-5)^{6}}$\n\nusing the chain rule for exponential…

Explanation:

Step1: Let $u=(x^{2}-5)^{6}$

$y = e^{u}$

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

Since $y = e^{u}$, $\frac{dy}{du}=e^{u}$

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

Using the chain - rule on $u=(x^{2}-5)^{6}$. Let $v = x^{2}-5$, then $u = v^{6}$. First, $\frac{du}{dv}=6v^{5}$ and $\frac{dv}{dx}=2x$. So, $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=6(x^{2}-5)^{5}\cdot2x = 12x(x^{2}-5)^{5}$

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $\frac{dy}{du}=e^{u}$ and $\frac{du}{dx}=12x(x^{2}-5)^{5}$ into the chain - rule formula. Since $u=(x^{2}-5)^{6}$, we have $\frac{dy}{dx}=e^{(x^{2}-5)^{6}}\cdot12x(x^{2}-5)^{5}=12x(x^{2}-5)^{5}e^{(x^{2}-5)^{6}}$

Answer:

$12x(x^{2}-5)^{5}e^{(x^{2}-5)^{6}}$