Q1 - Calculus

Question

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.

$$\begin{gathered} \left(a\right) True or False: \frac{d}{dx}\left(5\right)=0 \\ \left(b\right) True or False: \frac{d}{dr}(ks+r)=k \\ \left(c\right) True or False: \frac{d}{ds}(ks+r)=k \\ \left(d\right) True or False: \frac{d}{dx}{(3x+1)}^k=k{(3x+1)}^{k-1}\left(e\right) True or False: \frac{d}{dx}\left(\frac{1}{x^3}\right)=\frac{1}{3x^2} \\ \left(f\right) True or False : If f and g are differentiable functions, then \left(f\right(x\left)g\right(x\left)\right)\text{'}=g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)g\left(x\right) \\ \left(g\right) True or False : If f and g are differentiable functions, then \\ \left(\frac{g\left(x\right)}{h\left(x\right)}\right)\text{'}=\frac{h\left(x\right)g\text{'}\left(x\right)-g\left(x\right)h\text{'}\left(x\right)}{(h{\left(x\right))}^2} \\ \left(h\right)True or False: Proving the sum rule for differentiation involves the definition of the derivative , a lot of algebrac manipuations and the sum rule for limits \end{gathered}$$

Step-by-Step Solution

Answer

(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

 

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

$$\begin{gathered} \frac{d}{dr} (ks + r) = k \\ \frac{d}{dr} (ks + r) =\frac{d}{dr} ks +\frac{d}{dr} r=0+1=1 \\ \frac{d}{dr} (ks + r) = 1 \end{gathered}$$

The given statement is false  because the derivative is 1

Part(c): Step 1: Derivative

$$\begin{gathered} \frac{d}{ds} (ks + r) = k \\ \frac{d}{ds} (ks + r) = \frac{d}{ds}\left(ks\right)+\frac{d}{ds}\left(r\right)=k+0=k \end{gathered}$$

The given statement is true because derivative is k 

Part(d) : Step 1: Derivative

$$\begin{gathered} \frac{d}{dx}{(3x+1)}^k=k{(3x+1)}^{k-1} \\ \frac{d}{dx}{(3x+1)}^k=k{(3x+1)}^{k-1}*\left(3\right)=3k{(3x+1)}^{k-1} \end{gathered}$$

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

Part(e): Step 1 : Derivative

$$\begin{gathered} \mathbf{ }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{ }\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{)}\mathbf{\hspace{0.17em}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{(}\mathbf{1}\mathbf{/}{\mathit{x}}^{\mathbf{3}}\mathbf{)}\mathbf{ }\mathbf{=}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}{\mathit{x}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{=}\frac{\mathbf{-}\mathbf{3}\mathbf{ }}{{\mathbf{x}}^{\mathbf{3}}} \end{gathered}$$

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

Part f: step 1: Derivative

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

The given statement is true because it represents the product rule 

Part(g): Step 1 Derivative

If g and h are differentiable functions, then

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

The given statement is true because it represent the Quotient rule 

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 .