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Expert-verified Found in: Page 197 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.$\left(a\right)TrueorFalse:\frac{d}{dx}\left(5\right)=0\phantom{\rule{0ex}{0ex}}\left(b\right)TrueorFalse:\frac{d}{dr}\left(ks+r\right)=k\phantom{\rule{0ex}{0ex}}\left(c\right)TrueorFalse:\frac{d}{ds}\left(ks+r\right)=k\phantom{\rule{0ex}{0ex}}\left(d\right)TrueorFalse:\frac{d}{dx}{\left(3x+1\right)}^{k}=k{\left(3x+1\right)}^{k-1}\left(e\right)TrueorFalse:\frac{d}{dx}\left(\frac{1}{{x}^{3}}\right)=\frac{1}{3{x}^{2}}\phantom{\rule{0ex}{0ex}}\left(f\right)TrueorFalse:Iffandgaredifferentiablefunctions,then\left(f\left(x\right)g\left(x\right)\right)\text{'}=g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}\left(g\right)TrueorFalse:Iffandgaredifferentiablefunctions,then\phantom{\rule{0ex}{0ex}}\left(\frac{g\left(x\right)}{h\left(x\right)}\right)\text{'}=\frac{h\left(x\right)g\text{'}\left(x\right)-g\left(x\right)h\text{'}\left(x\right)}{\left(h{\left(x\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}\left(h\right)TrueorFalse:Provingthesumrulefordifferentiationinvolvesthedefinitionofthederivative,alotofalgebracmanipuationsandthesumruleforlimits$

(a) The given statement is true as derivative of constant is 0

(b) The given statement is false because the derivative is 1

(c) The given statement is true because derivative is k

(d) The given statement is false because the derivative of 3x+1 is not taken .

(e) The derivative is not same. the given statement is false.

(f) The given statement is true because it represent the product rule

(g) The given statement is true because it represent the Quotient rule

(h) The given statement is true because the sum rule of limit is used for proving the sum rule of differentiation

See the step by step solution

## step 1:Given Information

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

## Part (a): Step 1 : Derivative

Derivative of a constant is 0.

$\frac{d}{dx}\left(5\right)=0$

The given statement is true as derivative of constant is 0

## Part(b): Step 1: Derivative

$\frac{d}{dr}\left(ks+r\right)=k\phantom{\rule{0ex}{0ex}}\frac{d}{dr}\left(ks+r\right)=\frac{d}{dr}ks+\frac{d}{dr}r=0+1=1\phantom{\rule{0ex}{0ex}}\frac{d}{dr}\left(ks+r\right)=1$

The given statement is false because the derivative is 1

## Part(c): Step 1: Derivative

$\frac{d}{ds}\left(ks+r\right)=k\phantom{\rule{0ex}{0ex}}\frac{d}{ds}\left(ks+r\right)=\frac{d}{ds}\left(ks\right)+\frac{d}{ds}\left(r\right)=k+0=k$

The given statement is true because derivative is k

## Part(d) : Step 1: Derivative

$\frac{d}{dx}{\left(3x+1\right)}^{k}=k{\left(3x+1\right)}^{k-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}{\left(3x+1\right)}^{k}=k{\left(3x+1\right)}^{k-1}*\left(3\right)=3k{\left(3x+1\right)}^{k-1}$

The given statement is false because the derivative for 3x+1 is not multiplied.

## Part(e): Step 1 : Derivative

$\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{\right)}\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\mathbf{1}\mathbf{/}{\mathbit{x}}^{\mathbf{3}}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}{\mathbit{x}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathbit{x}}^{\mathbf{\left(}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{1}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{-}\mathbf{3}\mathbf{}}{{\mathbf{x}}^{\mathbf{3}}}$

The derivative is not same. the given statement is false.

## Part f: step 1: Derivative

$\left(f\left(x\right)g\left(x\right)\right)\text{'}=g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)g\left(x\right)$

The given statement is true because it represents the product rule

## Part(g): Step 1 Derivative

If g and h are differentiable functions, then

$\left(\frac{g\left(x\right)}{h\left(x\right)}\right)\text{'}=\frac{\left(h\left(x\right){g}^{\text{'}}\left(x\right)-g\left(x\right){h}^{\text{'}}\left(x\right)\right)}{\left(h\left(x{\right)}^{2}\right)}$

The given statement is true because it represent the Quotient rule

## Part(h): Step 1 : Derivative

Proving the sum rule for differentiation involves the definition of the derivative, a lot of algebraic manipulation, and the sum rule for limits.

The given statement is true because the sum rule of limit is used for proving the sum rule of differentiation . ### Want to see more solutions like these? 