question find the derivative of y = tan^(8x + 3)(x).

Explanation:

Step1: Let $u = 8x + 3$ and $y=\tan^{u}(x)$.

Let $y = v^{u}$ where $v=\tan(x)$.

Step2: Use the chain - rule and product - rule.

First, find $\frac{\partial y}{\partial v}$ and $\frac{\partial y}{\partial u}$ and $\frac{dv}{dx}$ and $\frac{du}{dx}$. $\frac{\partial y}{\partial v}=uv^{u - 1}=u\tan^{u - 1}(x)$ (power rule for $y = v^{u}$ with respect to $v$), $\frac{\partial y}{\partial u}=v^{u}\ln(v)=\tan^{u}(x)\ln(\tan(x))$ (derivative of $a^{x}=a^{x}\ln(a)$ with $a = \tan(x)$), $\frac{dv}{dx}=\sec^{2}(x)$ (derivative of $\tan(x)$) and $\frac{du}{dx}=8$.

Step3: Apply the chain - rule $\frac{dy}{dx}=\frac{\partial y}{\partial v}\frac{dv}{dx}+\frac{\partial y}{\partial u}\frac{du}{dx}$.

$\frac{dy}{dx}=(8x + 3)\tan^{8x+2}(x)\sec^{2}(x)+8\tan^{8x + 3}(x)\ln(\tan(x))$.

Answer:

$(8x + 3)\tan^{8x+2}(x)\sec^{2}(x)+8\tan^{8x + 3}(x)\ln(\tan(x))$