Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^\text{x}}}\times\text{x}^{\text{e}^{\text{x}}}\Big\{\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{e}^{\text{e}^\text{x}}\Big\{\frac{1}{\text{x}}+\text{e}^\text{x}\times\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^{\text{e}}}\times\text{x}^{\text{e}-1}\Big\{\text{x}+\text{e}\log\text{x}\Big\}$
✓

Answer

We have, $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\text{y}=\text{u}+\text{v}+\text{w}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}\ .....(\text{i})$
Where $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}},\text{v}=\text{x}^{\text{e}^{\text{e}^{\text{x}}}}\text{ and w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
Now, $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\ .....(\text{ii})$
Taking log on both sides,
$\log\text{u}=\log\text{e}^{\text{x}^{\text{e}^{\text{x}}}}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\log\text{e}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\ .....(\text{iii})$
Taking $\log$ on both sides,
$\log\log\text{u}=\log\text{x}^{\text{e}^\text{x}}$
$\Rightarrow\log\log\text{u}=\text{e}^\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\log\text{u}}\frac{\text{d}}{\text{dx}}(\log\text{u})=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{1}{\log\text{u}}\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\log\text{u}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{A})$
[Using equation (ii) and (iii)]
Now, $\text{v}=\text{x}^{\text{e}^{\text{e}^\text{x}}}\ .....(\text{iv})$
Taking log on both sides,
$\log\text{v}=\log\text{x}^{\text{e}^{\text{e}^\text{x}}}$
$\Rightarrow\log\text{v}=\text{e}^{\text{e}^\text{x}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{e}^\text{x}}\big)$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\text{e}^\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{B})$
[Using equation (4)]
Now, $\text{w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}\ .....(\text{v})$
Taking log on sides,
$\log\text{w}=\log\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^\text{e}}\log\text{e}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^{\text{e}}}\ .....(\text{vi})$
Taking log on both sides,
$\log\log\text{w}=\log\text{x}^{\text{x}^{\text{e}}}$
$\Rightarrow\log\log\text{w}=\text{x}^{\text{e}}\log\text{x}$
$\Rightarrow\frac{1}{\log\text{w}}\frac{\text{d}}{\text{dx}} (\log\text{w})=\text{x}^\text{e}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^\text{e})$
$\Rightarrow\frac{1}{\log\text{w}}\big(\frac{1}{\text{w}}\big)\frac{\text{dw}}{\text{dx}}=\text{x}^{\text{e}}\big(\frac{1}{\text{x}}\big)\log\text{xex}^{\text{e}-1}$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{w}\log\text{w}\big[\text{x}^{\text{e}-1}+\text{e}\log\text{xx}^{\text{e}-1}\big]$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})\ .....(\text{C})$
Using equation (A), (B) and (C) in equation (i), we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Differentiation→Differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\sin\text{x.}$
→
If the sum of lengths of hypotenuse and a side of a right-angled triangle is given, show that area of triangle is maximum, when the angle between them is $\frac{\pi }{3}$
→
Prove that:
$\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}=1+\text{a}^2+\text{b}^2+\text{c}^2$
→
Evaluate the following integrals:
$\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}$
→
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
→
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
→
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
→
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\},-1<\text{x}<1$
→
The function f(x) is befined as follows: $\text{f(x)}=\begin{cases}\text{x}^2+\text{ax}+\text{b},&0\leq\text{x}<2\\3\text{x}+2,&2\leq\text{x}\leq4\\2\text{ax}+5\text{b},&4<\text{x}\leq8\end{cases}$ if is continuous on [0, 8] find the value of a and b.
→
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
→