Differentiation by taking log MCQ Quiz in বাংলা - Objective Question with Answer for Differentiation by taking log - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept Used:

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

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

Chain Rule: \(\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)\)

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

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

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

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

\(\Rightarrow \displaystyle \frac{dy}{dx}=\frac{y}{x}\)

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

\(\dfrac {p}{x} + \dfrac {q}{y}\cdot \dfrac {dy}{dx} = \dfrac {(p + q)}{(x + y)} \left (1 + \dfrac {dy}{dx}\right )\)

\(\left \{\dfrac {p}{x} - \dfrac {p + q}{x + y}\right \} = \left \{\dfrac {p + q}{x + y} - \dfrac {q}{y}\right \} \dfrac {dy}{dx}\)

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

\(\Rightarrow \dfrac {(py - qx)}{x} = \dfrac {(py - qx)}{y}\cdot \dfrac {dy}{dx}\)

\(\Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x}\)

Concept:

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

Formula used :

• (ln x)' = \(1 \over x\)
• \((\sqrt x)' = {1 \over 2 \sqrt x}\)

Calculation:

Given,  y = x\(\sqrt x\),

Taking log on both sides,

 ln y = \(\sqrt x\) ln x

Differentiating both sides,

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

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

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

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

∴ The correct option is (5).

Concept:

• \(\rm d\over dx\)xn = nxn-1
• \(\rm d\over dx\)sin x = cos x
• \(\rm d\over dx\)cos x = -sin x
• \(\rm d\over dx\)ex = ex
• \(\rm d\over dx\)ln x = \(\rm1\over x\)
• \(\rm d\over dx\)(ax + b) = a
• \(\rm d\over dx\)tan x = sec2 x
• \(\rm d\over dx\)f(x)g(x) = f'(x)g(x) + f(x)g'(x)
• \(\rm d\over dx\) sin-1 x = \(\rm 1\over\sqrt{1 - x^2}\)
• \(\rm d\over dx\) tan-1 x = \(\rm 1\over1 + x^2\)

Calculation:

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

Taking log both sides, we get

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

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

Let f(x) = y

Differentiating with respect to x, we get

\(\rm {1\over y}{dy\over dx}\) = \(\rm \cos x\cdot [log(\log x)] \ +\ \sin x\cdot({\frac {1}{x\cdot \log x}})\)

\(\rm {dy\over dx}\) = y × \(\rm [\cos x\cdot [log(\log x)] \ +\ \sin x\cdot({\frac {1}{x\cdot \log x}})]\)

\(\rm {dy\over dx}\) = \(\rm \rm (\log x)^{\sin x} \cdot[\cos x\cdot [log(\log x)] \ + \sin x\cdot({\frac {1}{x\cdot \log x}})]\)

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 = \(\frac{\rm x}{1 + \rm log x}\)

Differentiating both sides , we get

\(\frac{\mathrm{d} \rm y}{\mathrm{d} x}\) =   \(\frac{1 +\rm log x - x × \frac{1}{x} }{(1 + log x)^{2}}\) 

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

= \(\frac{\rm log x }{(1 + log x)^{2}}\)

Concept:

Parametric Form:

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

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

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

\(\rm {1\over z}{dz\over dx}\) = -2 ln x\(\rm\left({1\over x}\right)\)

\(\rm {dz\over dx} = z\left[-2\ln x\over x\right]\)

\(\rm {dz\over dx} = x^{-\ln x}\left[-2\ln x\over x\right]\)

\(\rm {dz\over dx} = -2x^{-\ln x}\left[\ln x\over x\right]\)

Also y = \(\rm e^{x^{2}}\)

Taking log both sides, we get

ln y = ln \(\rm e^{x^{2}}\)        

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

ln y = x²                      (∵ ln e = 1)

Differentiating with respect to x

\(\rm {1\over y}{dy\over dx} = 2x\)

\(\rm {dy\over dx} = y[2x]\)

\(\rm {dy\over dx} = 2x e^{x^{2}}\)

The required result is

\(\rm dz\over dy\) = \(\rm {dz\over dx}\over{dy\over dx}\)

= \(\rm -2x^{-\ln x}\left[\ln x\over x\right]\over2xe^{x^{2}}\)

= \(\rm -x^{-\ln x}(\ln x)\over x^2e^{x^{2}}\)

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.