Q. 1 C. - Calculus

Question

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. 

(a) True or False: To find the derivative of \(sin x\) we have to use the definition of the derivative.

(b) True or False: To find the derivative of \(tan x\) we have to use the definition of the derivative.

(c) True or False: The derivative of \(\frac{x^{4}}{sinx}\) is \(\frac{4x^{3}}{cosx}\)

(d) True or False: If a function is algebraic, then so is its derivative. 

(e) True or False: If a function is transcendental, then so is its derivative.

(f) True or False: If \(f\) is a trigonometric function, then \(f'\) is also a trigonometric function.

(g) True or False: If \(f\) is an inverse trigonometric function, then \(f'\) is also an inverse trigonometric function. 

(h) True or False: If \(f\) is a hyperbolic function, then \(f'\) is also a hyperbolic function.

Step-by-Step Solution

Answer

Part (a). True

Part (b). False

Part (c). False

Part (d). True

Part (e). False

Part (f). True

Part (g). False

Part (h). True

Part (a) Step 1. Explanation

The definition of derivative is,

\(\frac{d}{dx}(f(x))=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

Consider a function , then its derivative can be calculated as

\(\frac{d}{dx}(f(x))=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

\(=\lim_{h\rightarrow 0}\frac{(sin(x)cos(h)_+sin(h)cosx)-sin(x)}{h}\)

\(=\lim_{h\rightarrow 0}\frac{sinx(cos(h)-1)-sin(h)cosx}{h}\)

\(=sinx\left ( \lim_{h\rightarrow 0}\frac{(cos(h)-1)}{h} \right )+cosx\left ( \lim_{h\rightarrow 0}\frac{sin(h)}{h} \right )\)

\(=sinx(0)+cosx(1)\)

\(=cosx\)

Therefore, the statement “To find the derivative of \(sinx\) we have to use the definition of the derivative.” is true.

Part b. Step 1. Explanation

We do not have to resort to the definition of the derivative in order to prove the formula for differentiating . Instead, we can use the quotient rule, the fact that \(tan x=\frac{sin x}{cos c}\).

That is,

\(\frac{d}{dx}(tan x)=\frac{d}{dx}\left ( \frac{sinx}{cosx} \right )\)

\(=\frac{cosx\frac{d}{dx}(sinx)-sinx\frac{d}{dx}(cosx)}{(cosx)^{2}}\)

\(=\frac{cosx(cosx)-sinx(-sinx)}{cos^{2}x}\)

\(=\frac{cos^{2}x+sin^{2}x}{cos^{2}x}\)

\(=\frac{1}{cos^{2}x}\)

\(=sec^{2}x\)

Therefore, the statement “To find the derivative of we have to use the definition of the derivative.” is false.

Part c. Step 1. Explanation

The derivative of \(\frac{x^{4}}{sinx}\) is,

\(\frac{d}{dx}\left ( \frac{x^{4}}{sinx} \right )=\frac{sinx\frac{d}{dx}(x^{4})-x^{4}\frac{d}{dx}(sinx)}{(sinx)^{2}}\)

\(=\frac{sinx\cdot (4x^{3})-x^{4}(cosx)}{sin^{2}x}\)

\(=\frac{4x^{3}sinx-x^{4}cosx}{sin^{2}x}\)

Therefore, the statement “The derivative of \(\frac{x^{4}}{sinx}\) is \(\frac{4x^{3}}{cosx}\)'' is false. 

Part d. Step 1. Explanation

Consider an algebraic function \(f(x)=2x^{3}+3x\)

Then, its derivative is

\(\frac{d}{dx}(f(x))=\frac{d}{dx}(2x^{3}+3x)\)

\(=6x^{2}+3\)

Therefore, the statement “If a function is algebraic, then so is its derivative.” is true.

Part e. Step 1. Explanation

Consider a transcendental function \(f(x)=\frac{x^{2}+1}{cosx}\)

Then, its derivative is

\(\frac{d}{dx}(f(x))=\frac{d}{dx}\left ( \frac{x^{2}+1}{cosx} \right )\)

\(f'(x)=\frac{cosx\frac{d}{dx}(x^{2}+1)-(x^{2}+1)\frac{s}{dx}(cosx)}{(cosx)^{2}}\)

\(f'(x)=\frac{cosx\cdot (2x)-(x^{2}+1)\cdot (-sinx)}{cos^{2}x}\)

\(f'(x)=2xsec x+(x^{2}+1)tan x sec x\)

Therefore, the statement “If a function is transcendental, then so is its derivative.” is false.

Part f. Step 1. Calculation

Consider a trigonometric function \(f(x)=sin x\).

Then, its derivative is

\(\frac{d}{dx}(f(x))=\frac{d}{dx}(sinx)\)

\(=cosx\)

Therefore, the statement “If a function is trigonometric, then so is its derivative.” is true.

Part g. Step 1. Explanation

Consider an inverse trigonometric function \(f(x)=sin^{-1}x\).

Then, its derivative is

\(\frac{d}{dx}(f(x))=\frac{d}{dx}(sin^{-1}x)\frac{d}{dx}(f(x))=\frac{d}{dx}(sin^{-1}x)\)

\(=\frac{1}{\sqrt{1-x^{2}}}\)

Therefore, the statement “If \(f\) is an inverse trigonometric function, then \(f'\) is also an inverse trigonometric function.” is false.

Part (h). Step 1. Explanation

The derivatives of hyperbolic functions are,

\(sinhx=\frac{e^{x}-e^{-x}}{2}\)

\(coshx=\frac{e^{x}-e^{-x}}{2}\)

\(\frac{d}{dx}(sinhx)=coshx\)

\(\frac{d}{dx}(coshx)=sinhx\)

\(\frac{d}{dx}(tanhx)=sech^{2}x\)

Therefore, the statement “If \(f\) is an hyperbolic function, then \(f'\) is also a hyperbolic function.” is true.