Question

find the derivative using the difference quotient. f(x)=3sqrt(x+5)

Answer 1

@dan815

Answer 2

@Daniellelovee

Answer 3

@Astrophysics

Answer 4

@phi

Answer 5

@mathstudent55

Answer 6

what do you mean by the difference quotient? do you mean the
lim h->0 of f(x+h) - f(x) / h ?

Answer 7

Is 3*√(x+5) or not?

Answer 8

The general difference quotient for a function \(f(x)\) is the ratio,
\[\frac{f(x+h)-f(x)}{h}\]so the derivative is given by
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]
With \(f(x)=\sqrt[3]{x+5}\) (I'm assuming you meant cube root, and not \(3\) times the square root of \(5\) - correct me if I'm wrong), you have
\[f'(x)=\lim_{h\to0}\frac{\sqrt[3]{x+h+5}-\sqrt[3]{x+5}}{h}\]
The trick with this kind of limit lies behind the numerator being a difference of two cube roots. Recall the formula for a difference of cubes:
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]If you were to let \(a=\sqrt[3]{x+h+5}\) and \(b=\sqrt[3]{x+5}\), then you can rewrite the function as
\[\frac{a-b}{h}\times\frac{a^2+ab+b^2}{a^2+ab+b^2}=\frac{a^3-b^3}{h(a^2+ab+b^2)}\]or, undoing the substitution, you have
\[f'(x)=\lim_{h\to0}\frac{\bigg(\sqrt[3]{x+h+5}\bigg)^3-\bigg(\sqrt[3]{x+5}\bigg)^3}{h\bigg(\bigg(\sqrt[3]{x+h+5}\bigg)^2+\sqrt[3]{(x+h+5)(x+5)}+\bigg(\sqrt[3]{x+5}\bigg)^2\bigg)}\]From here you can simplify.