Question

$\rm Given \\ \rm f(x,y) = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) } \\ \rm find \: f_{x}(1,0)$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x,y) = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) }$

So,

$\rm \: f(1,0) = 1( {1}^{2} + {0}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {1}^{2} \times 0) }$

$\rm \: f(1,0) = ( 1 + 0)^{ \frac{ - 3}{2} } {e}^{sin0}$

$\rm \: f(1,0) = 1 \times {e}^{0}$

$\rm\implies \:\rm \: f(1,0) = 1$

Now, Consider again

$\rm \: f(x,y) = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) }$

can be rewritten as

$\rm \: f = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) }$

On taking log on both sides, we get

$\rm \:log f = logx - \frac{3}{2}log( {x}^{2} + {y}^{2} ) + {sin( {x}^{2}y)}loge$

$\rm \:log f = logx - \frac{3}{2}log( {x}^{2} + {y}^{2} ) + {sin( {x}^{2}y)}$

On differentiating partially w. r. t. x, we get

$\rm \:\dfrac{\partial }{\partial x}log f = \dfrac{\partial }{\partial x}logx - \frac{3}{2}\dfrac{\partial }{\partial x}log( {x}^{2} + {y}^{2} ) + \dfrac{\partial }{\partial x}{sin( {x}^{2}y)}$

$\rm \: \frac{1}{f} \dfrac{\partial f}{\partial x}= \dfrac{1}{x} - \frac{3}{2} \times \dfrac{2x }{ {x}^{2} + {y}^{2} } + {cos( {x}^{2}y)}(2xy)$

$\rm \: \frac{1}{f} \dfrac{\partial f}{\partial x}= \dfrac{1}{x} - \dfrac{3x}{ {x}^{2} + {y}^{2} } + 2xy{cos( {x}^{2}y)}$

$\rm \: \dfrac{\partial f}{\partial x}= f\bigg(\dfrac{1}{x} - \dfrac{3x}{ {x}^{2} + {y}^{2} } + 2xy{cos( {x}^{2}y)} \bigg)$

$\rm\implies \:\rm \: f_x= f(x,y)\bigg(\dfrac{1}{x} - \dfrac{3x}{ {x}^{2} + {y}^{2} } + 2xy{cos( {x}^{2}y)} \bigg)$

Now, on substituting x = 1 and y = 0, we get

$\rm \: f_x(1,0)= f(1,0)\bigg(\dfrac{1}{1} - \dfrac{3 \times 1}{ {1}^{2} + {0}^{2} } + 2(1)(0){cos(0)} \bigg)$

$\rm \: f_x(1,0)= 1 \times \bigg(1 - 3 + 0 \bigg)$

$\rm\implies \:\rm \: f_x(1,0) \: = \: - \: 2$

$\rule{190pt}{2pt}$

Additional Information :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question:-

$\sf \large f(x,y) = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) } \\ \sf \large find \: f_{x}(1,0).$

Given:-

$\sf \large f(x,y) = x( {x}^{2} + {y}^{2} )^{ \frac{ - 3}{2} } {e}^{sin( {x}^{2}y) } .$

To Find:-

$\sf \large find \: f_{x}(1,0).$

Solution:-

$\sf \large f(x,y) = x(x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{2} } e {}^{sin(x {}^{2} y)}$

$\sf \large f_{x} = \frac{∂}{∂x} \: f(x,y) \\ \\ \sf \large y⇒constant.$

$\sf \large f_{x} = (x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{2} } e {}^{sin(x {}^{2}y) } \frac{∂}{∂x} (x) + x \: \: e {}^{sin(x {}^{2} y)} \frac{∂}{∂x} (x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{2} } + x(x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{2} } \frac{∂}{∂x} e {}^{sin(x {}^{2}y) }$

$\sf \large = (x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{ 2} } e {}^{sin(x {}^{2}y) } (1) + x \: e {}^{sin(x {}^{2}y) } \bigg( \frac{ - 3}{2} \bigg)(x {}^{2} + y {}^{2} ) {}^{ \frac{ - 5}{2} } (2x) + x(x {}^{2} + y {}^{2} ) {}^{ \frac{ - 3}{2} } e {}^{sin(x {}^{2}y) } \: cos(x {}^{2} y)(2x)(y)$

$\sf \large f_{x}(1,0)$

$\sf \large = (1) {}^{ \frac{ - 3}{2} } (1) + (1)(1) \bigg( \frac{ - 3}{2} \bigg)(1)(2) + (1)(1)(1)(1)(2)(0)$

$\sf \large = 1 - 3 + 0 = - 2$

Answer:-

${ \boxed{ \rm \huge \color{red}{f_{x}(1,0) = - 2.}}}$

Hope you have satisfied. ⚘