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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
Logarithmic differentiation is a powerful technique to differentiate functions of the form $\text{f}(\text{x})=[\text{u}(\text{x})]^{\text{v}(\text{x})},$ where both u(x) and v(x) are differentiable functions and f and u need to be positive functions. Let function $\text{y}=\text{f}(\text{x})=(\text{u}(\text{x}))^{\text{v}(\text{x})},$ then $\text{y}'=\text{y}\Big[\frac{\text{v}(\text{x})}{\text{u}(\text{x})}\text{u}'(\text{x})+\text{v}'(\text{x})\cdot\log[\text{u}(\text{x})]\Big]$ On the basis of above information, answer the following questions.
  1. Differentiate $x^x$ w.r.t. $x.$
  1. $\text{x}^\text{x}(1+\log\text{x})$
  2. $\text{x}^\text{x}(1-\log\text{x})$
  3. $-\text{x}^\text{x}(1+\log\text{x})$
  4. $\text{x}^\text{x}\log\text{x}$
  1. Differentiate $x^x + a^x+ x^a + a^a$ w.r.t. $x.$
  1. $(1+\log\text{x})+(\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1})$
  2. $\text{x}^\text{x}(1+\log\text{x})+\log\text{a}+\text{ax}^{\text{a}-1}$
  3. $\text{x}^\text{x}(1+\log\text{x})+\text{x}^\text{a}\log\text{x}+\text{ax}^{\text{a}-1}$
  4. $\text{x}^\text{x}(1+\log\text{x})+\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$
  1. If $\text{x}=\text{e}^\frac{\text{x}}{\text{y}},$ then find $\frac{\text{dy}}{\text{dx}}.$
  1. $-\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
  2. $-\frac{(\text{x}-\text{y})}{\text{x}\log\text{x}}$
  3. $\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
  4. $\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
  1. If $y = (2 - x)^3(3 + 2x)^5,$ then find $\frac{\text{dy}}{\text{dx}}.$
  1. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}-\frac{8}{2-\text{x}}\Big]$
  2. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
  3. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$
  4. $(2-\text{x})^3(3+2\text{x})^5\cdot\Big[\frac{10}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
  1. If $\text{y}=\text{x}^\text{x}\cdot\text{e}^{(2\text{x}+5)},$ then find $\frac{\text{dy}}{\text{dx}}.$
  1. $\text{x}^\text{x}\text{e}^{2\text{x}+5}$
  2. $\text{x}^\text{x}\text{e}^{2\text{x}+5}(3-\log\text{x})$
  3. $\text{x}^\text{x}\text{e}^{2\text{x}+5}(1-\log\text{x})$
  4. $\text{x}^\text{x}\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$
✓

Answer

  1. (a) $\text{x}^\text{x}(1+\log\text{x})$
Solution:
Let $\text{y}=\text{x}^\text{x}\Rightarrow\log\text{y}=\text{x}\log\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1\times\log\text{x}+\text{x}\times\frac{1}{\text{x}}]$
$=\text{x}^\text{x}[1+\log\text{x}]$
  1. (d) $\text{x}^\text{x}(1+\log\text{x})+\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$
  1. (d) $\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
Solution:
Given $\text{x}=\text{e}^\frac{\text{x}}{\text{y}}\Rightarrow\log\text{x}=\frac{\text{x}}{\text{y}}\log\text{e}\Rightarrow\text{y}\log\text{x}=\text{x}$
$\Rightarrow\text{y}\frac{1}{\text{x}}+(\log\text{x})\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(1-\frac{\text{y}}{\text{x}}\Big)\frac{1}{\log\text{x}}\Rightarrow\frac{1}{\text{x}\log\text{x}}(\text{x}-\text{y})$
  1. (c) $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$
Solution:
$\text{y}=(2-\text{x})^3(3+2\text{x})^5$
$\Rightarrow\log\text{y}=\log(2-\text{x})^3+\log(3+2\text{x})^5$
$=3\log(2-\text{x})+5\log(3+2\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{3\times(-1)}{2-\text{x}}+\frac{5}{3+2\text{x}}\times(2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$
  1. (d) $\text{x}^\text{x}\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$
Solution:
$\text{y}=\text{x}^\text{x}\cdot\text{e}^{(2\text{x}+5)}$
$\Rightarrow\log\text{y}=\text{x}\log\text{x}+(2\text{x}+5)$
$\Rightarrow\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\Big(\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\Big)+2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cdot\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$

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