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Q6:

Differentiate the functions w.r.t. $x$. $\left(x + \frac{1}{x}\right)^x + x^{\left(1 + \frac{1}{x}\right)}$

Solution :

Step 1: Decomposition of the Function

Let the given function be $y = f(x) + g(x)$, where:

  • $f(x) = \left(x + \frac{1}{x}\right)^x$
  • $g(x) = x^{\left(1 + \frac{1}{x}\right)}$

To find the derivative $\frac{dy}{dx}$, we must compute $\frac{df}{dx}$ and $\frac{dg}{dx}$ separately using logarithmic differentiation [Per the property $\frac{d}{dx}[u^v] = u^v \left( \frac{v}{u} \cdot \frac{du}{dx} + \ln(u) \cdot \frac{dv}{dx} \right)$].

Step 2: Differentiating $f(x) = \left(x + \frac{1}{x}\right)^x$

Taking the natural logarithm on both sides:

$\ln(f) = x \ln\left(x + \frac{1}{x}\right) = x \ln\left(\frac{x^2 + 1}{x}\right) = x [\ln(x^2 + 1) - \ln(x)]$

Differentiating with respect to $x$ using the Product Rule [$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$]:

$\frac{1}{f} \frac{df}{dx} = (x) \cdot \frac{d}{dx}[\ln(x^2 + 1) - \ln(x)] + [\ln(x^2 + 1) - \ln(x)] \cdot \frac{d}{dx}(x)$

$\frac{1}{f} \frac{df}{dx} = x \left( \frac{2x}{x^2 + 1} - \frac{1}{x} \right) + \ln\left(\frac{x^2 + 1}{x}\right)$

$\frac{df}{dx} = f(x) \left[ \frac{2x^2}{x^2 + 1} - 1 + \ln\left(x + \frac{1}{x}\right) \right] = \left(x + \frac{1}{x}\right)^x \left[ \frac{x^2 - 1}{x^2 + 1} + \ln\left(x + \frac{1}{x}\right) \right]$

Step 3: Differentiating $g(x) = x^{\left(1 + \frac{1}{x}\right)}$

Taking the natural logarithm on both sides:

$\ln(g) = \left(1 + \frac{1}{x}\right) \ln(x)$

Differentiating with respect to $x$:

$\frac{1}{g} \frac{dg}{dx} = \left(1 + \frac{1}{x}\right) \cdot \frac{d}{dx}(\ln x) + \ln(x) \cdot \frac{d}{dx}\left(1 + \frac{1}{x}\right)$

$\frac{1}{g} \frac{dg}{dx} = \left(\frac{x + 1}{x}\right) \cdot \frac{1}{x} + \ln(x) \cdot \left(-\frac{1}{x^2}\right)$

$\frac{dg}{dx} = g(x) \left[ \frac{x + 1}{x^2} - \frac{\ln x}{x^2} \right] = x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 - \ln x}{x^2} \right]$

Step 4: Combining Results

Since $y = f(x) + g(x)$, then $\frac{dy}{dx} = \frac{df}{dx} + \frac{dg}{dx}$.

Term Result
$\frac{df}{dx}$ $\left(x + \frac{1}{x}\right)^x \left[ \frac{x^2 - 1}{x^2 + 1} + \ln\left(x + \frac{1}{x}\right) \right]$
$\frac{dg}{dx}$ $x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 - \ln x}{x^2} \right]$

Final Solution:

$\frac{dy}{dx} = \left(x + \frac{1}{x}\right)^x \left[ \frac{x^2 - 1}{x^2 + 1} + \ln\left(x + \frac{1}{x}\right) \right] + x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 - \ln x}{x^2} \right]$


More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.5


CBSE Solutions for Class 12 Mathematics Continuity and Differentiability


Chapters in CBSE - Class 12 Mathematics


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