Find d y d x , if y =x^ x^x - Vidyadip

Question

Find $\frac{ d y}{ d x}$, if $y =x^{x^x}$

✓

Answer

$y=x^{x^x}$
Taking logarithm of both sides, we get
$\log y=\log x^{x^x}$
$\therefore \log y = x ^{ x } \log x$
Differentiating both sides w.r.t. $x$, we get
$ \frac{ d }{ d x}(\log y)=x^x \cdot \frac{ d }{ d x}(\log x)+\log x \cdot \frac{ d }{ d x}\left(x^x\right)$
$\therefore \frac{1}{y} \cdot \frac{ dy}{ d x}=x^x \cdot \frac{1}{x}+\log x \cdot \frac{ d }{ d x}\left(x^x\right) \ldots . . \text { (i) } $
Let $u = x ^{ x }$
Taking logarithm of both sides, we get
$\log u=\log x^x$
$\therefore \log u=x \log x$
Differentiating both sides w.r.t. $x$, we get
$ \frac{ d }{ d x}(\log u )=x \cdot \frac{ d }{ d x}(\log x)+\log x \cdot \frac{ d }{ d x}(x)$
$\therefore \frac{1}{ u } \cdot \frac{ du }{ d x}=x \cdot \frac{1}{x}+\log x \cdot 1$
$\therefore \frac{1}{ u } \cdot \frac{ du }{ d x}=1+\log x$
$\therefore \frac{ du }{ d x}= u (1+\log x )$
$\therefore \frac{ d }{ d x}\left(x^x\right)= x ^{ x }(1+\log x ) \quad \ldots . . . \text { (ii) } $
Substituting (ii) in (i), we get
$ \frac{1}{y} \cdot \frac{ d y}{ d x}=x^x \cdot \frac{1}{x}+\log x \cdot x^x(1+\log x)$
$\therefore \frac{ d y}{ d x}=y x^x\left[\frac{1}{x}+\log x(1+\log x)\right] $
$\therefore \frac{ d y}{ d x}=x^{x^x} \cdot x^x\left[\frac{1}{x}+\log x(1+\log x)\right]$

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 "Solve the Following Question.(4 Marks)" from Question Bank [2022]→Question Bank [2022] — all question types→Maths (commerce) STD 12 Commerce / Arts question papers→Maharashtra Board STD 12 Commerce / Arts question papers→Maharashtra Board question paper generator→

Similar questions

Find the area of the circle $x^2 + y^2 = 16$

Fit a trend line to the data in Problem IV (1) by the method of least squares.

If a $r.v. x.$ has $p.d.f f(x)=\left\{\begin{array}{l}\frac{c}{x}, 10 \\ 0, \text { otherwise }\end{array}\right.$ Find $c, E(X), $ and $\operatorname{Var}(X)$. Also Find $F(x)$.

Solve the following differential equation $\frac{ d y}{ d x}=\cos ( x + y )$
Solution: $\frac{ d y}{ d x}=\cos ( x + y )$$\ldots(i)$
Put
$ \therefore 1+\frac{ d y}{ d x}=\frac{ dv }{ d x}$
$\therefore \frac{ d y}{ d x}=\frac{ dv }{ d x}-1 $
$\therefore$ (1) becomes $\frac{ dv }{ d x}-1=\cos v$
$\therefore \frac{ dv }{ d x}=1+\cos v$
$\therefore \square dv = dx$
Integrating, we get
$ \int \frac{1}{1+\cos v } d v=\int d x$
$\therefore \int \frac{1}{2 \cos ^2\left(\frac{ v }{2}\right)} dv =\int d x$
$\therefore \frac{1}{2} \int \square dv =\int d x$
$\therefore \frac{1}{2} \cdot \frac{\tan \left(\frac{ v }{2}\right)}{\frac{1}{2}}= x + c$
$\therefore \square= x + c $

Calculate Walsh’s price Index Number for the following data.

Commodity	Base Year		Current Year
Commodity	Price	Quantity	Price	Quantity
I	10	12	40	3
II	20	2	25	8
III	30	3	50	27
IV	60	9	90	36

Write the dual statement of each of the following compound statements:
(i) 13 is prime number and India is a democratic country.
(ii) Karina is very good or everybody likes her.
(iii) Radha and Sushmita can not read Urdu.
(iv) A number is real number and the square of the number is non-negative.

A bill was drawn on $14$th April for $\text{₹} 7,000$ and was discounted on $6$th July at $5\%$ p.a. The Banker paid $\text{₹} 6,930$ for the bill. Find the period of the bill.

The equations of the two lines of regression are $6x + y − 31 = 0$ and $3x + 2y – 26 = 0.$ Find the value of the correlation coefficient

Find values $x$ and $y$ if the Price Index Number by Simple Aggregate Method by taking $2001$ as base year is $120 ,$ given $\sum p_1=300$.

Commodity	A	B	C	D
Price (in ₹) in 2001	$90$	$x$	$90$	$30$
Price (in ₹) in 2004	$95$	$60$	$y$	$35$

Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.

x	1	2	3
P(X = x)	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

$\text { Solution: } \mu=E(X)=\sum_{i=1}^3 x_i p_i$
$E(X)=\square+\square+\square=\square$
$\operatorname{Var}(X)=E\left(X^2\right)-\{E(X)\}^2$
$=\sum X_i^2 P_i-\left[\sum X_i P_i\right]^2$
$=\square-\square$
$=\square$