Question

1. differentiate:

a) $\frac{d}{dx}(2)=$
b) $\frac{d}{dx}(5^{0})=$
c) $\frac{d}{dx}(x)=$
d) $\frac{d}{dx}((x^{2})^{3})=$
e) $\frac{d}{dx}((\frac{1}{x^{2}})^{3})=$
f) $\frac{d}{dx}(x^{3}sqrt{x^{2}})=$
g) $\frac{d}{dx}(2x^{3})=$
h) $\frac{d}{dx}((\frac{3}{x})^{3})=$
i) $\frac{d}{dx}(8sqrt{x})=$

Explanation:

Step1: Recall derivative of constant

The derivative of a constant $c$ with respect to $x$ is 0. Since 2 is a constant, $\frac{d}{dx}(2)=0$.

Step2: Simplify and apply constant - derivative rule

First, $5^{0}=1$ (any non - zero number to the power of 0 is 1). Then, since 1 is a constant, $\frac{d}{dx}(5^{0})=\frac{d}{dx}(1) = 0$.

Step3: Apply power - rule for $y = x$

The power - rule for differentiation is $\frac{d}{dx}(x^{n})=nx^{n - 1}$. For $y=x=x^{1}$, using the power - rule, $\frac{d}{dx}(x)=1\times x^{1 - 1}=1$.

Step4: Simplify and apply power - rule

Simplify $(x^{2})^{3}=x^{6}$ (using the power - of - a - power rule $(a^{m})^{n}=a^{mn}$). Then, by the power - rule $\frac{d}{dx}(x^{6})=6x^{6 - 1}=6x^{5}$.

Step5: Simplify and apply power - rule

Simplify $(\frac{1}{x^{2}})^{3}=\frac{1}{x^{6}}=x^{-6}$. Then, by the power - rule $\frac{d}{dx}(x^{-6})=-6x^{-6 - 1}=-6x^{-7}=-\frac{6}{x^{7}}$.

Step6: Simplify and apply power - rule

Simplify $x\sqrt[3]{x^{2}}=x\times x^{\frac{2}{3}}=x^{1+\frac{2}{3}}=x^{\frac{5}{3}}$. Then, by the power - rule $\frac{d}{dx}(x^{\frac{5}{3}})=\frac{5}{3}x^{\frac{5}{3}-1}=\frac{5}{3}x^{\frac{2}{3}}=\frac{5}{3}\sqrt[3]{x^{2}}$.

Step7: Apply constant - multiple and power rules

By the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$ and the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $y = 2x^{3}$, $\frac{d}{dx}(2x^{3})=2\frac{d}{dx}(x^{3})=2\times3x^{3 - 1}=6x^{2}$.

Step8: Simplify and apply power - rule

Simplify $(\frac{3}{x})^{3}=\frac{27}{x^{3}}=27x^{-3}$. Then, by the power - rule $\frac{d}{dx}(27x^{-3})=27\times(-3)x^{-3 - 1}=-81x^{-4}=-\frac{81}{x^{4}}$.

Step9: Apply constant - multiple and power rules

Rewrite $8\sqrt{x}=8x^{\frac{1}{2}}$. Then, by the constant - multiple rule $\frac{d}{dx}(8x^{\frac{1}{2}})=8\frac{d}{dx}(x^{\frac{1}{2}})$. By the power - rule $\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}$. So, $\frac{d}{dx}(8\sqrt{x})=8\times\frac{1}{2}x^{-\frac{1}{2}} = 4x^{-\frac{1}{2}}=\frac{4}{\sqrt{x}}$.

Answer:

a) 0
b) 0
c) 1
d) $6x^{5}$
e) $-\frac{6}{x^{7}}$
f) $\frac{5}{3}\sqrt[3]{x^{2}}$
g) $6x^{2}$
h) $-\frac{81}{x^{4}}$
i) $\frac{4}{\sqrt{x}}$