Question

let $y(x)=b^{x}+x^{b}$. determine $y(x)$. assume that $b$ is a positive constant.

Explanation:

Step1: Differentiate $b^x$

The derivative of $a^x$ with respect to $x$ is $a^x\ln(a)$. So, the derivative of $b^x$ with respect to $x$ is $b^x\ln(b)$.

Step2: Differentiate $x^b$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, the derivative of $x^b$ with respect to $x$ is $bx^{b - 1}$.

Step3: Use sum - rule of differentiation

If $y = u+v$, then $y'=u'+v'$. Here $u = b^x$ and $v = x^b$. So $y'(x)=b^x\ln(b)+bx^{b - 1}$.

Answer:

$b^x\ln(b)+bx^{b - 1}$