Differentiation by taking log MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiation by taking log - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept Used:

Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

Product Rule: $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$

Chain Rule: $\frac{d}{dx}f(g(x))=f^{\prime }(g(x))g^{\prime }(x)$

Logarithmic Differentiation

Calculation

Given: $\{\frac{d}{dx}(x^x+x^{x+1}+x^{x+2}){\}}_{x=e}$

⇒ $\frac{d}{dx}(x^x+x^{x+1}+x^{x+2})=\frac{d}{dx}(x^x(1+x+x^2))$

Let $y=x^x$. ⇒ $\ln y=x\ln x$.

⇒ $\frac{1}{y}\frac{dy}{dx}=\ln x+x\cdot \frac{1}{x}=\ln x+1$.

⇒ $\frac{dy}{dx}=y(\ln x+1)=x^x(\ln x+1)$.

Using product rule:

⇒ $\frac{d}{dx}(x^x(1+x+x^2))=x^x(\ln x+1)(1+x+x^2)+x^x(1+2x)$

At $x=e$:

⇒ $e^e(\ln e+1)(1+e+e^2)+e^e(1+2e)$

⇒ $e^e(1+1)(1+e+e^2)+e^e(1+2e)$

⇒ $2e^e(1+e+e^2)+e^e(1+2e)$

⇒ $e^e(2+2e+2e^2+1+2e)$

⇒ $e^e(2e^2+4e+3)$

∴ The value is $e^e(2e^2+4e+3)$.

Hence option 3 is correct.

Taking $\log$ on both sides, we get

$p\log x+q\log y=(p+q)\log (x+y)$

$\Rightarrow {\displaystyle \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{(p+q)}{(x+y)}(1+\frac{dy}{dx})}$

$\Rightarrow {\displaystyle (\frac{p}{x}-\frac{p+q}{x+y})=(\frac{p+q}{x+y}-\frac{q}{y})\frac{dy}{dx}}$

$\Rightarrow {\displaystyle \frac{dy}{dx}=\frac{y}{x}}$

If $x^p+y^q=(x+y{)}^{p+q}$

Taking log on both sides,

$p\log x+q\log y=(p+q)\log (x+y)$

On differentiating w.r.t. x, we get

${\displaystyle \frac{p}{x}}+{\displaystyle \frac{q}{y}}\cdot {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{(p+q)}{(x+y)}}(1+{\displaystyle \frac{dy}{dx}})$

$\{{\displaystyle \frac{p}{x}}-{\displaystyle \frac{p+q}{x+y}}\}=\{{\displaystyle \frac{p+q}{x+y}}-{\displaystyle \frac{q}{y}}\}{\displaystyle \frac{dy}{dx}}$

$\left\{{\displaystyle \frac{px+py-px-qx}{x(x+y)}}\right\}=\left\{{\displaystyle \frac{py+qy-qx-qy}{y(x+y)}}\right\}{\displaystyle \frac{dy}{dx}}$

$\Rightarrow {\displaystyle \frac{(py-qx)}{x}}={\displaystyle \frac{(py-qx)}{y}}\cdot {\displaystyle \frac{dy}{dx}}$

$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$

Concept:

Product rule: (f.g)'(x) = f '(x).g(x) + f(x).g'(x)

Formula used :

• (ln x)' = $\frac{1}{x}$
• $(\sqrt{x}{)}^{\prime }=\frac{1}{2\sqrt{x}}$

Calculation:

Given,  y = x$\sqrt{x}$,

Taking log on both sides,

 ln y = $\sqrt{x}$ ln x

Differentiating both sides,

⇒ $\frac{1}{y}.\frac{dy}{dx}=\frac{1}{2\sqrt{x}}.\ln x+\sqrt{x}.\frac{1}{x}$

⇒ $\frac{1}{y}.\frac{dy}{dx}=\frac{1}{2\sqrt{x}}.\ln x+\frac{1}{\sqrt{x}}$

⇒ $\frac{1}{y}.\frac{dy}{dx}=\frac{(\ln x+2)}{2\sqrt{x}}$

⇒ $\frac{dy}{dx}=\frac{y.(\ln x+2)}{2\sqrt{x}}$

∴ The correct option is (5).

Concept:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$xn = nxn-1
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$sin x = cos x
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$cos x = -sin x
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$ex = ex
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$ln x = $\frac{1}{\mathrm{x}}$
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$(ax + b) = a
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$tan x = sec2 x
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$f(x)g(x) = f'(x)g(x) + f(x)g'(x)
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$ sin-1 x = $\frac{1}{\sqrt{1-{\mathrm{x}}^2}}$
• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}$ tan-1 x = $\frac{1}{1+{\mathrm{x}}^2}$

Calculation:

Given: f(x) = $(\log \mathrm{x}{)}^{\sin \mathrm{x}}$

Taking log both sides, we get

⇒ log f(x) = log $(\log \mathrm{x}{)}^{\sin \mathrm{x}}$

⇒ log f(x) = sin x [log(log x)]               (∵ log mn = n log m)

Let f(x) = y

Differentiating with respect to x, we get

$\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = $\cos \mathrm{x}\cdot [\mathrm{l}\mathrm{o}\mathrm{g}(\log \mathrm{x})]+\sin \mathrm{x}\cdot (\frac{1}{\mathrm{x}\cdot \log \mathrm{x}})$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = y × $[\cos \mathrm{x}\cdot [\mathrm{l}\mathrm{o}\mathrm{g}(\log \mathrm{x})]+\sin \mathrm{x}\cdot (\frac{1}{\mathrm{x}\cdot \log \mathrm{x}})]$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = $(\log \mathrm{x}{)}^{\sin \mathrm{x}}\cdot [\cos \mathrm{x}\cdot [\mathrm{l}\mathrm{o}\mathrm{g}(\log \mathrm{x})]+\sin \mathrm{x}\cdot (\frac{1}{\mathrm{x}\cdot \log \mathrm{x}})]$

Calculation:

x^y = e^{x - y} 

Taking log on both sides, we get 

⇒ xy = log ex - y 

⇒ y log x = (x - y) log_e e 

⇒ y log x = x - y             [∵ loge e = 1]

⇒ ( 1 + log x)y = x  

⇒ y = $\frac{\mathrm{x}}{1+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}}$

Differentiating both sides , we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}x}$ =   $\frac{1+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}-\mathrm{x}\times \frac{1}{\mathrm{x}}}{(1+logx{)}^2}$ 

= $\frac{1+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}-1}{(1+logx{)}^2}$ 

= $\frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}}{(1+logx{)}^2}$

Concept:

Parametric Form:

If f(x) and g(x) are the functions in x, then 

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}$ 

Calculation:

Let z = x^{-ln x} 

Taking log both sides, we get

ln z = ln x-ln x

ln z = - (ln x)²                       (∵ ln mⁿ = n ln m)

Differentiating with respect to x

$\frac{1}{\mathrm{z}}\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}$ = -2 ln x$\left(\frac{1}{\mathrm{x}}\right)$

$\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}=\mathrm{z}\left[\frac{-2\ln \mathrm{x}}{\mathrm{x}}\right]$

$\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}={\mathrm{x}}^{-\ln \mathrm{x}}\left[\frac{-2\ln \mathrm{x}}{\mathrm{x}}\right]$

$\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}=-2{\mathrm{x}}^{-\ln \mathrm{x}}\left[\frac{\ln \mathrm{x}}{\mathrm{x}}\right]$

Also y = ${\mathrm{e}}^{{\mathrm{x}}^2}$

Taking log both sides, we get

ln y = ln ${\mathrm{e}}^{{\mathrm{x}}^2}$        

ln y = x² ln e               (∵ ln mn = n ln m)

ln y = x²                      (∵ ln e = 1)

Differentiating with respect to x

$\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2\mathrm{x}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{y}[2\mathrm{x}]$

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

The required result is

$\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{y}}$ = $\frac{\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$

= $\frac{-2{\mathrm{x}}^{-\ln \mathrm{x}}\left[\frac{\ln \mathrm{x}}{\mathrm{x}}\right]}{2\mathrm{x}{\mathrm{e}}^{{\mathrm{x}}^2}}$

= $\frac{-{\mathrm{x}}^{-\ln \mathrm{x}}(\ln \mathrm{x})}{{\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2}}$

Concept:

log₁₀(e) = 1 and

(log₁₀(x))^y = ylog(x)

Calculation:

Given:

$x=e^{y+e^{y+e^{y+...}}}$

For infinite terms, the above equation can be written as:

$x=e^{y+x}$

Taking log both sides

log(x) = (y + x)loge

log(x) = y + x

Differentiating both sides:

$\frac{1}{x}=\frac{dy}{dx}+1$

$\frac{dy}{dx}=\frac{1-x}{x}$

Hence option (2) is the solution.