Second Order Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Second Order Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculations:

Given, f(x) = ecos x

Taking derivative with respect to x on both side, we get

⇒ f'(x) = ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}$(ecos x)

⇒ f'(x) = ecos x ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}(\cos \mathrm{x})$

⇒ f'(x) = ecos x $(-\sin \mathrm{x})$

⇒ f'(x) = $(-\sin \mathrm{x})$ecos x                                     (∴ e^cosx = f(x))

⇒ f''(x) = $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$(- sin x)f(x)

⇒ f''(x) = - sin x.f'(x) + f(x)(- cos x)

⇒ f''(x) = - sin x.f'(x) - f(x).cos x

Putting the value of f(x) and f'(x)

⇒ f''(x) = sin² x e^{cos x} - cos x ecos x

⇒ f''(x) = ecos x(sin² x - cos x)

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

Formula:

$\frac{\mathrm{d}{\mathrm{e}}^{\mathrm{x}}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}$

Calculation:

Let y = e^2x                          ..... (1)

Differentiating with respect to x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}({\mathrm{e}}^{2\mathrm{x}})$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$  = 2e2x

Again, differentiating with respect to x, we get

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

From equation (1), we get

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

Concept:

Second-order Derivative.

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=$ $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{x})]$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{e}}^{\mathrm{f}(\mathrm{x})}={\mathrm{e}}^{\mathrm{f}(\mathrm{x})}{\mathrm{f}}^{\prime }(\mathrm{x})$

Calculations:

y = e5x 

Differentiating w.r. to x on both side, we get

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}=5{\mathrm{e}}^{5\mathrm{x}}$

Again Differentiating w.r. to x on both sides, we get

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}=5\times 5{\mathrm{e}}^{5\mathrm{x}}$

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}=25\mathrm{y}$

Concept:

Chain Rule of Derivatives:

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

Derivatives of Trigonometric Functions:

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

Calculation:

x = a cos t

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}}=\mathrm{a}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}}\cos \mathrm{t}=-\mathrm{a}\sin \mathrm{t}$

y = b sin t

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}=\mathrm{b}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}}\sin \mathrm{t}=\mathrm{b}\cos \mathrm{t}$

Now, ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}\times {\displaystyle \frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}}$

= $\mathrm{b}\cos \mathrm{t}\times {\displaystyle \frac{1}{-\mathrm{a}\sin \mathrm{t}}}$

= $-{\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\times \cot \mathrm{t}$

And, ${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}}={\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\left({\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}\right)$

= $(-{\displaystyle \frac{\mathrm{b}}{\mathrm{a}}})({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\cot \mathrm{t})$

= $-\left({\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\right)\times ({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}}\cot \mathrm{t})\times {\displaystyle \frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}}$

= $(-{\displaystyle \frac{\mathrm{b}}{\mathrm{a}}})\times (-{\csc }^2\mathrm{t})\times {\displaystyle \frac{1}{-\mathrm{a}\sin \mathrm{t}}}$

= ${\displaystyle \frac{\mathrm{b}}{{\mathrm{a}}^2}}\times {\displaystyle \frac{-1}{{\sin }^3\mathrm{t}}}$

= ${\displaystyle \frac{\mathrm{b}}{{\mathrm{a}}^2}}\times {\displaystyle \frac{-1}{{\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}^3}}$

= ${\displaystyle \frac{-\mathrm{b}}{{\mathrm{a}}^2}}\times {\left({\displaystyle \frac{\mathrm{b}}{\mathrm{y}}}\right)}^3$

= $-{\displaystyle \frac{{\mathrm{b}}^4}{{\mathrm{a}}^2{\mathrm{y}}^3}}$.

Concept:

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

Calculation:

Let y = x³ - x² + 1.

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

y' = 3x² - 2x.

Differentiating again (double derivative) w.r.t. x, we get:

y'' = 6x - 2.

Concept:

Second-order Derivative.

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=$ $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{x})]$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{e}}^{\mathrm{f}(\mathrm{x})}={\mathrm{e}}^{\mathrm{f}(\mathrm{x})}{\mathrm{f}}^{\prime }(\mathrm{x})$

Calculations:

Let y = ekx 

Differentiating w.r. to x on both side, we get

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}=\mathrm{k}{\mathrm{e}}^{\mathrm{k}\mathrm{x}}$

Again Differentiating w.r. to x on both sides, we get

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}={\mathrm{k}}^2{\mathrm{e}}^{\mathrm{k}\mathrm{x}}$

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}={\mathrm{k}}^2\mathrm{y}$

Hence, If y = ekx then ${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}={\mathrm{k}}^2\mathrm{y}$ 

Concept:

Second-order Derivative.

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=$ $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{x})]$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x}$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}$

Calculations:

 ${\displaystyle \frac{{\mathrm{d}}^2\sin 2\mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}}$

= ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\sin 2\mathrm{x})$

= ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}(2\mathrm{c}\mathrm{o}\mathrm{s}2\mathrm{x})$

= $2{\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}(\mathrm{c}\mathrm{o}\mathrm{s}2\mathrm{x})$

= $2(-2\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x})$

= -4sin 2x

Hence, ${\displaystyle \frac{{\mathrm{d}}^2\sin 2\mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}}$ = - 4sin 2x

Calculation:

Given:

y = 5 cos x − 3 sin x 

Differentiating both sides w.r.t x

$\frac{dy}{dx}$= −5 sin x − 3 cos x 

Again differentiating both sides w.r.t x

 $\frac{d^{2}y}{d{x}^2}$ = −5 cos x + 3 sin x

= −y

Concept:

Second-order Derivative.

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=$ $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{x})]$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{e}}^{\mathrm{f}(\mathrm{x})}={\mathrm{e}}^{\mathrm{f}(\mathrm{x})}{\mathrm{f}}^{\prime }(\mathrm{x})$

Calculations:

y = x3 - 6x2​ + 2 + sin x

Differentiating w.r. to x on both side, we get

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$ = ( 3 x2) - (6 × 2 x) + 0 + cos x

⇒ 3x2 - 12x + cos x

Again Differentiating w.r. to x on both sides, we get 

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}$ = (3 × 2 x) - 12 - sin x

⇒${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d}\mathrm{x}}^2}}=6\mathrm{x}-12-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}$

Concept: 

If y = f(x) and x = f(t)  then by chain Rule $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}.\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}$

Calculations:

Given, y = sin (ax + b)

Take derivative on both sides,we get 

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{a}\mathrm{x}+\mathrm{b})).\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\mathrm{a}\mathrm{x}+\mathrm{b})$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{a}\mathrm{x}+\mathrm{b}).\mathrm{a}$

Again take derivative on both sides, we get$$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}.\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\mathrm{a}.\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{a}\mathrm{x}+\mathrm{b})]$

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=-{\mathrm{a}}^2\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{a}\mathrm{x}+\mathrm{b})$

Now, to find  ${\displaystyle \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}}$ at $\mathrm{x}=-\frac{\mathrm{b}}{\mathrm{a}}$, put $\mathrm{x}=-\frac{\mathrm{b}}{\mathrm{a}}$in above equation.
$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=-{\mathrm{a}}^2\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{a}\times \frac{-\mathrm{b}}{\mathrm{a}}+\mathrm{b})$
$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=-{\mathrm{a}}^2\mathrm{s}\mathrm{i}\mathrm{n}(0)$
$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=0$