How do you differentiate #g(w) = (w^n - 5^n) /n #?

Question

Christopher Agresta · Accepted Answer

Assuming that #n# is a constant, #g'(w) = w^(n-1)# Assuming that #n# is a constant, #g(w) = (w^n - 5^n) /n =1/n(w^n-5^n)# and #5^n# is constant, so its derivative is #0#. #g'(w) = 1/n d/(dw)(w^n-5^n)# # = 1/n (nw^(n-1)-0) = w^(n-1)#

Ezra Adams · Answer

undefined To differentiate the function $g(w) = \frac{w^n - 5^n}{n}$, we'll apply the rules of differentiation. Notice that $5^n$ is a constant, and $n$ in the denominator is also a constant with respect to $w$. Thus, the differentiation process mainly concerns the $w^n$ term.

The differentiation of $g(w)$ with respect to $w$ is performed as follows:

$$ g(w) = \frac{w^n - 5^n}{n} $$

$$ g'(w) = \frac{d}{dw} \left( \frac{w^n}{n} \right) - \frac{d}{dw} \left( \frac{5^n}{n} \right) $$

Since $\frac{5^n}{n}$ is a constant with respect to $w$, its derivative is 0:

$$ g'(w) = \frac{d}{dw} \left( \frac{w^n}{n} \right) - 0 $$

$$ g'(w) = \frac{1}{n} \cdot \frac{d}{dw}(w^n) $$

Using the power rule for differentiation, where $\frac{d}{dw}(w^n) = nw^{n-1}$, we get:

$$ g'(w) = \frac{1}{n} \cdot nw^{n-1} $$

$$ g'(w) = w^{n-1} $$

So, the derivative of $g(w) = \frac{w^n - 5^n}{n}$ with respect to $w$ is $g'(w) = w^{n-1}$.