Application of Derivatives

$f\left(x\right)=e^{x^{2}ln\left(\frac{2}{x}\right)}=e^{x^2\left(ln2-lnx\right)}$

$f\text{'}\left(x\right)=e^{x^2\left(ln2-lnx\right)}\left[2x\left(ln2-lnx\right)-x\right]$            

$={\left(\frac{2}{x}\right)}^{x^2}x\left[2\left(ln2-lnx\right)-1\right]$          

$f\text{'}\left(x\right)=0\Rightarrow 2\left(ln2-lnx\right)-1=0$   

2ln x = $ln\frac{4}{e}\Rightarrow x^2=\frac{4}{e}$  

$f\left(x\right)=2x^3-3x^2-12x$  

  $f\text{'}\left(x\right)=6x^2-6x-12$             

= 6 (x – 2) (x + 1)

a = -1, b = 2

$A={\displaystyle \underset{-1}{\overset{0}{\int }}\left(2x^3-3x^2-12x\right)dx-{\displaystyle \underset{0}{\overset{2}{\int }}\left(2x^3-3x^2-12x\right)dx=\frac{57}{2}}}$

->4A = 114

$f\left(x\right)=\left|\left(x-1\right)\left(x+1\right)\left(x-3\right)\right|+x-3$  

For    $x\in \left[-1,1\right]\cup \left[3,\infty \right)$

y = x² (x – 3)

  $\frac{dy}{dx}=2x\left(x-3\right)+x^2=3x^2-6x=3x\left(x-2\right)$

Number of max = 2

Number of min = 1

$S=\pi r\mathcal{l}$

$=\pi x\sqrt[]{x^2+y^2}$

$=\pi x\sqrt[]{x^2+25x^2}$

$\frac{ds}{dt}=\pi \sqrt[]{26}.2x.\frac{dx}{dt}$  

${\left(\frac{ds}{dt}\right)}_{y=10}\to x=2$                

$5\pi .3x^2\frac{dx}{dt}=1$                

$\left(\frac{dx}{dt}\right)=\frac{1}{15\pi \left(4\right)}=\frac{1}{60\pi }$

$f\left(x\right)=y=a{x}^3+b{x}^2+cx+5$ ……. (i)

$\frac{dy}{dx}=3a{x}^2+2bx+c$ ……. (ii)

Touches x-axis at P (-2, 0)

$\Rightarrow y{|}_{x=-2}=0\Rightarrow -8a+4b-2c+5=0$ ……… (iii)

Touches x –axis at P (-2, 0) also implies

$\frac{dy}{dx}{|}_{x=-2}=0\Rightarrow 12a-4b+c=0$ ……… (iv)

y = f (x) cuts y-axis at (0, 5)

Given,

$\frac{dy}{dx}{|}_{x=0}=c=3$ ……. (v)

From (iii), (iv) and (v)

f (x) = 0 at x = -2 and x = 1

Local maximum value of f (x) is at x = 1

i.e., $\frac{27}{4}$

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ - 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$        

 For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$  which does not satisfy the equation so no horizontal

For vertical tangent -> $5y^4-9x=0$  

->m = 0, N = 2

  $CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

$=\frac{1}{2}*CD*AB=\frac{1}{2}*2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$              

 3x² = 10

x = k

3k² = 10

$f\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt}$

$f\text{'}\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt+e^x.\frac{f\text{'}\left(x\right)}{e^x}-\left[\left(2x-1\right).e^x+\left(x^2-x+1\right).e^x\right]}$

$\begin{array}{l}f\left(x\right)=\left(2x+1\right).e^x-2e^x+c\\ f\left(x\right)=e^x\left(2x-1\right)c=0\end{array}$

$f\left(-\frac{1}{2}\right)=\frac{-2}{\sqrt[]{e}}$

$\frac{5x^2}{2}+\frac{\alpha }{2x^5}+\frac{\alpha }{2x^5}\ge 7{\left(\frac{{\alpha }^2}{2^7}\right)}^{\frac{1}{7}}$

$\frac{7.{\left(\alpha \right)}^{2/7}}{2}=14$

${\left({\alpha }^2\right)}^{\frac{1}{7}}=2^2\Rightarrow \alpha ={\left(2^2\right)}^{\frac{7}{2}}=2^7$

=128

$\left|z-3\sqrt[]{2}\right|+\left|z-p\sqrt[]{2}i\right|$ is minimum for z,  $3\sqrt[]{2}\&p\sqrt[]{2}i$ are collinear.

$\iff {\left(3\sqrt[]{2}\right)}^2+{\left(p\sqrt[]{2}\right)}^2={\left(5\sqrt[]{2}\right)}^2$

$\Rightarrow 18+2p^2=50$

$2p^2=32$

$p=\pm 4$