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Q1

Expert-verifiedFound in: Page 197

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.

$\left(a\right)TrueorFalse:\frac{d}{dx}\left(5\right)=0\phantom{\rule{0ex}{0ex}}\left(b\right)TrueorFalse:\frac{d}{dr}(ks+r)=k\phantom{\rule{0ex}{0ex}}\left(c\right)TrueorFalse:\frac{d}{ds}(ks+r)=k\phantom{\rule{0ex}{0ex}}\left(d\right)TrueorFalse:\frac{d}{dx}{(3x+1)}^{k}=k{(3x+1)}^{k-1}\left(e\right)TrueorFalse:\frac{d}{dx}\left(\frac{1}{{x}^{3}}\right)=\frac{1}{3{x}^{2}}\phantom{\rule{0ex}{0ex}}\left(f\right)TrueorFalse:Iffandgaredifferentiablefunctions,then\left(f\right(x\left)g\right(x\left)\right)\text{'}=g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}\left(g\right)TrueorFalse:Iffandgaredifferentiablefunctions,then\phantom{\rule{0ex}{0ex}}\left(\frac{g\left(x\right)}{h\left(x\right)}\right)\text{'}=\frac{h\left(x\right)g\text{'}\left(x\right)-g\left(x\right)h\text{'}\left(x\right)}{(h{\left(x\right))}^{2}}\phantom{\rule{0ex}{0ex}}\left(h\right)TrueorFalse:Provingthesumrulefordifferentiationinvolvesthedefinitionofthederivative,alotofalgebracmanipuationsandthesumruleforlimits$

(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

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.

Derivative of a constant is 0.

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

The given statement is true as derivative of constant is 0

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

The given statement is false because the derivative is 1

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

The given statement is true because derivative is k

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

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

$\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{)}\mathbf{\hspace{0.17em}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{(}\mathbf{1}\mathbf{/}{\mathit{x}}^{\mathbf{3}}\mathbf{)}\mathbf{}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}{\mathit{x}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{=}\frac{\mathbf{-}\mathbf{3}\mathbf{}}{{\mathbf{x}}^{\mathbf{3}}}$

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

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

The given statement is true because it represents the product rule

If g and h are differentiable functions, then

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

The given statement is true because it represent the Quotient rule

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 .

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