Question

1. differentiate each of the following.

a) $f(x)=x^{3}+5x + 2$
b) $y=\frac{2 + 3x}{4-5x}$
c) $g(x)=\frac{x^{4}-7x^{3}+sqrt{x}}{x^{2}}$

Explanation:

Step1: Apply power - rule to \(f(x)\)

The power - rule states that if \(y = x^n\), then \(y^\prime=nx^{n - 1}\), and the derivative of a constant times a function is the constant times the derivative of the function, and the derivative of a constant is 0. For \(f(x)=x^{3}+5x + 2\), we have \(f^\prime(x)=\frac{d}{dx}(x^{3})+\frac{d}{dx}(5x)+\frac{d}{dx}(2)\).
\[f^\prime(x)=3x^{2}+5+0 = 3x^{2}+5\]

Step2: Apply quotient - rule to \(y\)

The quotient - rule states that if \(y=\frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). For \(y = \frac{2 + 3x}{4-5x}\), let \(u = 2 + 3x\), \(u^\prime=3\), \(v = 4-5x\), \(v^\prime=-5\). Then \(y^\prime=\frac{3(4 - 5x)-(2 + 3x)(-5)}{(4 - 5x)^{2}}=\frac{12-15x + 10 + 15x}{(4 - 5x)^{2}}=\frac{22}{(4 - 5x)^{2}}\)

Step3: Simplify \(g(x)\) first

\[g(x)=\frac{x^{4}-7x^{3}+\sqrt{x}}{x^{2}}=x^{2}-7x+x^{-\frac{3}{2}}\]
Then apply the power - rule. \(g^\prime(x)=\frac{d}{dx}(x^{2})-\frac{d}{dx}(7x)+\frac{d}{dx}(x^{-\frac{3}{2}})\)
\[g^\prime(x)=2x-7-\frac{3}{2}x^{-\frac{5}{2}}\]

Answer:

a) \(f^\prime(x)=3x^{2}+5\)
b) \(y^\prime=\frac{22}{(4 - 5x)^{2}}\)
c) \(g^\prime(x)=2x - 7-\frac{3}{2x^{\frac{5}{2}}}\)