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

Calculation:

⇒ $\mathrm{y}=\frac{1-{\mathrm{x}}^{32}}{1-\mathrm{x}}\Rightarrow \mathrm{y}-\mathrm{x}\mathrm{y}=1-{\mathrm{x}}^{32}$

⇒ y' - xy' - y = -32x³¹

⇒ y" - xy" - y' - y' = -(32)(31)x³⁰

at x = – 1 ⇒ y' - y" = 496

Hence, the correct answer is Option 3. 

Calculation: 

Applying C₃ → C₃ -C₁

$[y(x)=\left|\begin{array}{ccc}\sin x& \cos x& 1+\cos x-\sin x\\ 27& 28& 27-27\\ 1& 1& 1-1\end{array}\right|=\left|\begin{array}{ccc}\sin x& \cos x& 1+\cos x-\sin x\\ 27& 28& 0\\ 1& 1& 0\end{array}\right|$

$y(x)=(\sin x\cdot 0-\cos x\cdot 0+(1+\cos x-\sin x)(27-28))$

$y(x)=(1+\cos x-\sin x)(-1)=-1-\cos x+\sin x$

$\frac{dy}{dx}=-(-\sin x)+\cos x=\sin x+\cos x$

$\frac{d^{2}y}{d{x}^2}=\cos x-\sin x$

$\frac{d^{2}y}{d{x}^2}+y=(\cos x-\sin x)+(-1-\cos x+\sin x)=-1$

Hence, the correct answer is Option 1.

Concept:

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

• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

Formulas:

$\frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$

Calculation:

$\frac{{\mathrm{d}}^2\log \mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}$

= $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}$

= $\frac{\mathrm{d}\left(\frac{1}{\mathrm{x}}\right)}{\mathrm{d}\mathrm{x}}$

= $\frac{-1}{{\mathrm{x}}^2}$

Calculation:

 y = 100e2x - 200e-2x

Differentiate y with respect to x:

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

Differentiate again with respect to x:

⇒ $\frac{d^{2}y}{d{x}^2}=400e^{2x}-800e^{-2x}$

Factor out 4:

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

⇒ $\frac{d^{2}y}{d{x}^2}=4y$

Comparing with $\frac{d^{2}y}{d{x}^2}=ay$, 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

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

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 $y^{″}+b^{2}y$, we get

⇒ $y^{″}+b^{2}y=(-a^2\sin ax-b^2\cos bx)+b^2(\sin ax+\cos bx)$

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

⇒ $(b^2-a^2)\sin ax$

∴ $y^{″}+b^{2}y=(b^2-a^2)\sin ax$

Hence option 1 is correct

Concept:

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

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

Formulas:

$\frac{\mathrm{d}{\cot }^{-1}\mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{-1}{1+{\mathrm{x}}^2}$

Calculation:

$\frac{{\mathrm{d}}^2{\cot }^{-1}\mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}$

= $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}{\cot }^{-1}\mathrm{x}}{\mathrm{d}\mathrm{x}}$

= $\frac{d}{dx}(\frac{-1}{1+x^2})$

= $-\frac{d}{dx}(\frac{1}{1+x^2})$

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

⇒$-[\frac{-2\mathrm{x}}{(1+{\mathrm{x}}^2{)}^2}]$

= $\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2{)}^2}$

Concept:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)\pm \mathrm{g}\left(\mathrm{x}\right)\right\}=\frac{\mathrm{d}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}{\mathrm{d}\mathrm{x}}\pm \frac{\mathrm{d}\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}{\mathrm{d}\mathrm{x}}$
• $\frac{\mathrm{d}\left(\sin \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=\cos \mathrm{x}$
• $\frac{\mathrm{d}\left(\cos \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=-\sin \mathrm{x}$

Calculation:

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

Differentiating with respect to x, we get

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

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\left(\mathrm{p}\cos 2\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}+\frac{\mathrm{d}\left(\mathrm{q}\sin 2\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-2\mathrm{p}\sin 2\mathrm{x}+2\mathrm{q}\cos 2\mathrm{x}$

Again, differentiating with respect to x, we get

${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}}=-2\mathrm{p}\times \frac{\mathrm{d}\left(\sin 2\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}+2\mathrm{q}\times \frac{\mathrm{d}\left(\cos 2\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}$

$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}}=-4\mathrm{p}\cos 2\mathrm{x}-4\mathrm{q}\sin 2\mathrm{x} \\ =-4(\mathrm{p}\cos 2\mathrm{x}+\mathrm{q}\sin 2\mathrm{x}) \end{gathered}$$

From equation (1), we get

$\therefore {\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}}=-4\mathrm{y}$

Concept:

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.
• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$.

Product Rule of Derivatives:

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

Calculation:

We have x = $\frac{1-\mathrm{t}}{1+\mathrm{t}}$ and y = $\frac{2\mathrm{t}}{1+\mathrm{t}}$.

We observe that x + y = $\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right)+\left(\frac{2\mathrm{t}}{1+\mathrm{t}}\right)=\frac{1+\mathrm{t}}{1+\mathrm{t}}$ = 1.

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

1 + $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 0

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = -1

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

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=0$.

CONCEPT:

$\frac{\mathrm{d}\left(\sin \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=\cos \mathrm{x}$

$\frac{\mathrm{d}\left(\cos \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=-\sin \mathrm{x}$

$\frac{\mathrm{d}\left(\ln \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}},\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{x}>0$

CALCULATION:

Given: y = sin (log x)

First let's find out y₁

As we know that, $\frac{\mathrm{d}\left(\sin \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=\cos \mathrm{x}$ and $\frac{\mathrm{d}\left(\ln \mathrm{x}\right)}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}},\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{x}>0$

⇒ ${\mathrm{y}}_1=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})}{\mathrm{x}}$

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

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

⇒ ${\mathrm{y}}_2=\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{-\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})-\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})}{{\mathrm{x}}^2}$

Now, x²y₂ + xy₁ = -y

Hence, correct option is 2.

Concept:

$\frac{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{d}\mathrm{x}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

To Find: $\frac{{\mathrm{d}}^2({\mathrm{x}}^{20})}{\mathrm{d}{\mathrm{x}}^2}$

$\frac{{\mathrm{d}}^2({\mathrm{x}}^{20})}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}{\mathrm{x}}^{20}}{\mathrm{d}\mathrm{x}}\right)$

$=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(20{\mathrm{x}}^{19})=20\frac{\mathrm{d}{\mathrm{x}}^{19}}{\mathrm{d}\mathrm{x}}$

= 20 × 19 × x¹⁸

= 380x18 

Concept:

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

• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

Formulas:

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}({\tan }^{-1}\mathrm{x})=\frac{1}{1+{\mathrm{x}}^2}$

Calculation:

$\frac{{\mathrm{d}}^2}{\mathrm{d}{\mathrm{x}}^2}({\tan }^{-1}\mathrm{x})$

= $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}({\tan }^{-1}\mathrm{x})$

= $\frac{\mathrm{d}\left(\frac{1}{1+{\mathrm{x}}^2}\right)}{\mathrm{d}\mathrm{x}}$

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

= $\frac{-2\mathrm{x}}{(1+{\mathrm{x}}^2{)}^2}$

Concept:

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.
• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$.

Derivatives of Trigonometric Functions:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\tan \mathrm{x}={\sec }^2\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cot \mathrm{x}=-{\csc }^2\mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\csc \mathrm{x}=-\cot \mathrm{x}\csc \mathrm{x} \end{gathered}$$

Calculation:

Using the chain rule of derivatives, we get:

$\frac{\mathrm{d}\tan 2\mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\tan 2\mathrm{x}}{\mathrm{d}(2\mathrm{x})}\times \frac{\mathrm{d}(2\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = 2 sec² 2x

Again differentiation in respect to x, we get

$\frac{{\mathrm{d}}^2\tan 2\mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}\tan 2\mathrm{x}}{\mathrm{d}\mathrm{x}}$

= $\frac{2\mathrm{d}{\sec }^{2}2\mathrm{x}}{\mathrm{d}\mathrm{x}}$

=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: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

Formulas:

$\frac{\mathrm{d}\tan \mathrm{x}}{\mathrm{d}\mathrm{x}}={\sec }^2\mathrm{x}$

$\frac{\mathrm{d}\sec \mathrm{x}}{\mathrm{d}\mathrm{x}}=\sec \mathrm{x}\tan \mathrm{x}$

$\frac{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{d}\mathrm{x}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

We have to find the value of $\frac{{\mathrm{d}}^2\tan \mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}$

$\frac{{\mathrm{d}}^2\tan \mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}\tan \mathrm{x}}{\mathrm{d}\mathrm{x}}\right)$

$=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\right)$

Apply chain rule, we get

$=\frac{\mathrm{d}{\sec }^2\mathrm{x}}{\mathrm{d}\sec \mathrm{x}}\times \frac{\mathrm{d}\sec \mathrm{x}}{\mathrm{d}\mathrm{x}}$

= 2sec x . sec x tan x

= 2sec² x tan x

Concept:

$\frac{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{d}\mathrm{x}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

To Find: $\frac{{\mathrm{d}}^2({\mathrm{x}}^{10})}{\mathrm{d}{\mathrm{x}}^2}$

$\frac{{\mathrm{d}}^2({\mathrm{x}}^{10})}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}{\mathrm{x}}^{10}}{\mathrm{d}\mathrm{x}}\right)$

$=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(10{\mathrm{x}}^9)=10\frac{\mathrm{d}{\mathrm{x}}^9}{\mathrm{d}\mathrm{x}}$

= 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,

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-5cosx+3sinx=-y$

Hence, correct option is 4.