Second Order Derivatives MCQ Quiz - Objective Question with Answer for Second Order Derivatives - Download Free PDF

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm \frac{d \log x}{dx} = \frac{1}{x}\)

Calculation:

\(\rm \frac {d^2 \log x}{dx^2}\)

= \(\rm \frac{d}{dx} \times \frac {d\log x}{dx}\)

= \(\rm \frac{d \left(\frac {1}{x} \right )}{dx} \)

= \(\rm \frac{-1}{x^2}\)

Calculation:

 y = 100e2x - 200e-2x

Differentiate y with respect to x:

⇒ \(\frac{dy}{dx} = 200e^{2x} + 400e^{-2x}\)

Differentiate again with respect to x:

⇒ \(\frac{d^2y}{dx^2} = 400e^{2x} - 800e^{-2x}\)

Factor out 4:

⇒ \(\frac{d^2y}{dx^2} = 4(100e^{2x} - 200e^{-2x})\)

⇒ \(\frac{d^2y}{dx^2} = 4y\)

Comparing with \(\frac{d^2y}{dx^2} = ay\), we get a = 4.

Hence option 1 is correct

Formula Used:

Chain rule for differentiation

Calculation:

Given:

\(y = \sin ax + \cos bx\)

Differentiating both sides w.r.t. x, we get

⇒ \(y' = \frac{d}{dx}(\sin ax + \cos bx) = a\cos ax - b\sin bx\)

Differentiating again w.r.t. x, we get

⇒ \(y'' = \frac{d}{dx}(a\cos ax - b\sin bx) = -a^2\sin ax - b^2\cos bx\)

Now, substituting the values of y and y'' in the expression \(y'' + b^2y\), we get

⇒ \(y'' + b^2y = (-a^2\sin ax - b^2\cos bx) + b^2(\sin ax + \cos bx)\)

⇒ \( -a^2\sin ax - b^2\cos bx + b^2\sin ax + b^2\cos bx\)

⇒ \( (b^2 - a^2)\sin ax\)

∴ \(y'' + b^2y = (b^2 - a^2)\sin ax\)

Hence option 1 is correct

\((x-h)^2 + (y-h)^2 = r^2 \)

Differentiating and dividing by 2,

\(x - h + yy' - hy' = 0 \)

\(\Rightarrow h = \dfrac{x+yy'}{1+y'}\)

\(h(1 + y') = x + yy'\)

Again difference wrt x,

\(hy'' = 1 + yy'' + y'^2\)

Putting value of h,

\(\left(\dfrac{x + yy'}{1 + y'}\right)y' = 1 + yy'' + y'^2\)

\((x + yy')y'' = (1 + yy'' + y'^2) (1 + y')\)

\(\therefore\,\, xy'' + yy'y'' = 1 + yy'' + y'^2 + y' + yy''y' + y'^3\)

\(\therefore\,\, xy'' = 1 + yy'' + y'^2 + y' + y'^3\)

\(1 + yy'' + y'^2 + y'^3 + yy'' - xy'' = 0\)

\(1 + y' + y'^2 + y'^3 + yy'' - xy'' = 0\)

\(\therefore\,\,1 + y' (1 + y' + y'^2) + y'' (y - x) = 0\)

\(\Rightarrow P=y-x, Q=1+y'+(y')^{2}\)

\(\Rightarrow P+Q=1-x+y+y'+(y')^{2}\)

Calculation

We have ,

\((x-h)^2 + (y-h)^2 = r^2 \)

Differentiating and dividing by 2,

\(x - h + yy' - hy' = 0 \)

\(\Rightarrow h = \dfrac{x+yy'}{1+y'}\)

\(h(1 + y') = x + yy'\)

Again difference wrt x,

\(hy'' = 1 + yy'' + y'^2\)

Putting value of h,

\(\left(\dfrac{x + yy'}{1 + y'}\right)y' = 1 + yy'' + y'^2\)

\((x + yy')y'' = (1 + yy'' + y'^2) (1 + y')\)

\(\therefore\,\, xy'' + yy'y'' = 1 + yy'' + y'^2 + y' + yy''y' + y'^3\)

\(\therefore\,\, xy'' = 1 + yy'' + y'^2 + y' + y'^3\)

\(1 + yy'' + y'^2 + y'^3 + yy'' - xy'' = 0\)

\(1 + y' + y'^2 + y'^3 + yy'' - xy'' = 0\)

\(\therefore\,\,1 + y' (1 + y' + y'^2) + y'' (y - x) = 0\)

\(\Rightarrow P=y-x, Q=1+y'+(y')^{2}\)

\(\Rightarrow P+Q=1-x+y+y'+(y')^{2}\)

Hence option 2 and 3 are correct

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
• Division rule: \(\frac{d}{dx}\frac{u}{v}=\frac{v\times u'-u \times v'}{v^2}\)

Formulas:

\(\rm \frac{d \cot^{-1} x}{dx} = \frac{-1}{1+x^2}\)

Calculation:

\(\rm \frac {d^2 \cot^ {-1}x}{dx^2}\)

= \(\rm \frac{d}{dx} \times \frac {d\cot^{-1} x}{dx}\)

= \(\frac{d}{dx}(\frac{-1}{1+x^2})\)

= \(-\:\frac{d}{dx}(\frac{1}{1+x^2})\)

⇒ \(-[\frac{(1+x^2)\times 0-1 \times2x}{(1+x^2)^2}]\)

⇒\(-[\rm \frac{-2x}{(1+x^2)^2}] \)

= \(\rm \frac{2x}{(1+x^2)^2} \)

Concept:

• \(\rm \frac{d}{{dx}}\left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \frac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}} \pm \frac{{d\left\{ {g\left( x \right)} \right\}}}{{dx}}\)
• \(\rm \frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\rm \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)

Calculation:

Given: y = p cos 2x + q sin 2x                    .... (1)

Differentiating with respect to x, we get

As we know that, \(\rm \frac{d}{{dx}}\left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \frac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}} \pm \frac{{d\left\{ {g\left( x \right)} \right\}}}{{dx}}\)

\(\rm \Rightarrow \frac{{dy}}{{dx}} = \frac{{d\left( {p\cos 2x} \right)}}{{dx}} + \frac{{d\left( {q\sin 2x} \right)}}{{dx}}\)

\(\rm \Rightarrow \frac{{dy}}{{dx}} =-2p\sin2x+2q\cos2x\)

Again, differentiating with respect to x, we get

\(\rm \dfrac{d^2y}{dx^2}=-2p \times \frac{{d\left( {\sin 2x} \right)}}{{dx}} +2q \times \frac{{d\left( {\cos 2x} \right)}}{{dx}}\)

\(\rm \Rightarrow \dfrac{d^2y}{dx^2}=-4p\cos 2x - 4q\sin2x\\=-4(p\cos2x+q\sin2x)\)

From equation (1), we get

\(\rm \therefore \dfrac{d^2y}{dx^2}= -4y\)

Concept:

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

Product Rule of Derivatives:

• \(\rm \frac{d}{dx}[f(x)\times g(x)]=f(x)\times\frac{d}{dx}g(x)+\frac{d}{dx}f(x)\times g(x)\).

Calculation:

We have x = \(\rm {{1\ -\ t}\over {1\ +\ t}}\) and y = \(\rm {2t\over {1\ +\ t}}\).

We observe that x + y = \(\rm \left({{1\ -\ t}\over {1\ +\ t}}\right)+\left({{2t}\over {1\ +\ t}}\right)={1\ +\ t\over1\ + \ t}\) = 1.

Differentiating w.r.t. x, we get:

1 + \(\rm {dy\over dx}\) = 0

⇒ \(\rm {dy\over dx}\) = -1

Differentiating again w.r.t. x, we get:

\(\rm \frac{d^2y}{dx^2}=0\).

CONCEPT:

\(\rm \frac{{d\left( {\sin x} \right)}}{{dx}} = \; \cos x\)

\(\rm \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)

\(\rm \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

CALCULATION:

Given: y = sin (log x)

First let's find out y₁

As we know that, \(\rm \frac{{d\left( {\sin x} \right)}}{{dx}} = \; \cos x\) and \(\rm \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

⇒ \(\rm y_1 = \frac{dy}{dx} = \frac{cos\ (log x)}{x}\)

Now, again by differentiating the above equation with respect to x we get,

As we know that, \(\rm \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)

⇒ \(\rm y_2 =\frac{d^2y}{dx^2} = \frac{-sin \ (log \ x) \ - \ cos \ (log \ x)}{x^2}\)

Now, x²y₂ + xy₁ = -y

Hence, correct option is 2.

Concept:

\(\rm \frac{dx^{n}}{dx}= nx^{n-1}\)

Calculation:

To Find: \(\rm \frac{d^2 (x^{20})}{dx^2}\)

\(\rm \frac{d^2 (x^{20})}{dx^2} = \frac{d}{dx} \left(\frac{dx^{20}}{dx} \right )\)

\(\rm = \frac{d}{dx} (20x^{19})= 20 \frac{dx^{19}}{dx}\)

= 20 × 19 × x¹⁸

= 380x18 

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm \frac{d }{dx} (\tan^{-1} x)= \frac{1}{1+x^2}\)

Calculation:

\(\rm \frac {d^2}{dx^2}( \tan^ {-1}x)\)

= \(\rm \frac{d}{dx} \times \frac {d}{dx}(\tan^{-1} x)\)

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

= \(\rm \frac{-1}{(1+x^2)^2} \times (0 + 2x)\)

= \(\rm \frac{-2x}{(1+x^2)^2} \)

Concept:

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

Derivatives of Trigonometric Functions:

\(\rm \frac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cos x=-\sin x\\ \frac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cot x=-\csc^2 x\\ \frac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \frac{d}{dx}\csc x=-\cot x\csc x\)

Calculation:

Using the chain rule of derivatives, we get:

\(\rm \frac{d\tan 2x}{dx} =\rm \frac{d\tan 2x}{d (2x)} \times \frac {d(2x)}{dx}\) = 2 sec² 2x

Again differentiation in respect to x, we get

\(\rm \frac{d^2\tan 2x}{dx^2} = \frac d {dx} \times \frac {d \tan 2x}{dx}\)

= \(\rm \frac {2d\sec^2 2x}{dx}\)

=2 (2 sec 2x)(tan 2x sec 2x)(2)

= 8 tan 2x sec² 2x

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm\frac{d\tan x}{dx} = \sec^2 x\)

\(\rm\frac{d\sec x}{dx} =\sec x \tan x\)

\(\rm \frac{dx^{n}}{dx}= nx^{n-1}\)

Calculation:

We have to find the value of \(\rm \frac{d^2\tan x}{dx^2}\)

\(\rm \frac{d^2\tan x}{dx^2} = \frac{d}{dx} \left(\frac{d\tan x}{dx} \right )\)

\(\rm = \frac{d}{dx} \left(sec^2 x \right )\)

Apply chain rule, we get

\(\rm = \frac{d\sec^2 x}{d\sec x} × \frac{d\sec x}{dx}\)

= 2sec x . sec x tan x

= 2sec² x tan x

Concept:

\(\rm \frac{dx^{n}}{dx}= nx^{n-1}\)

Calculation:

To Find: \(\rm \frac{d^2 (x^{10})}{dx^2}\)

\(\rm \frac{d^2 (x^{10})}{dx^2} = \frac{d}{dx} \left(\frac{dx^{10}}{dx} \right )\)

\(\rm = \frac{d}{dx} (10x^{9})= 10 \frac{dx^{9}}{dx}\)

= 10 × 9 × x8

= 90(x8)

CONCEPT:

• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\frac{{d\left( {\cos x} \right)}}{{dx}} = -\sin x\)

CALCULATION:

Given: y = 5 cos x - 3 sin x

First let's find out dy/dx

As we know that, \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\) and \(\frac{{d\left( {\cos x} \right)}}{{dx}} = -\sin x\)

⇒ dy/dx = - 5 sin x - 3 cos x

Now, again by differentiating the above equation with respect to x we get,

\(\Rightarrow \frac{d^2y}{dx^2} = - 5 \ cos \ x + 3 \ sin \ x = - y\)

Hence, correct option is 4.