Question

Please answer this question​

Answer 1

We are given with the function
`{\bf{f(x,y,z)=e^(x+y)\cos (z)+(y+1)\sin^(-1)(x)at}}` and we need to find
`{\bf \\abla f}` , that's nothing but just the gradient of f(x,y,z) . But before starting let's recall ;

For a function F(x, y, z, ....) , the gradient is given by ;

• 
  `{\boxed{\bf{\\abla F=(\partial F)/(\partial x)\hat{i}+(\partial F)/(\partial y)\hat{j}+(\partial F)/(\partial z)\hat{k}+\cdots}}}`

So , now let's calculate the partial derivatives of f(x, y, z) first with respect to x , y and z one after other .So consider ;

`{:\implies \quad \sf f(x,y,z)=e^(x+y)\cos (z)+(y+1)\sin^(-1)(x)at}`

Partial differentiating both sides w.r.t.x will yield ;

`{:\implies \quad \sf (\partial f)/(\partial x)=e^(x+y)\cos (z)+(y+1)at\frac{1}{\sqrt{1-x^(2)}}}`

Simplifying will yield ;

`{:\implies \quad \sf (\partial f)/(\partial x)=\frac{e^(x+y)\cos (z)\sqrt{1-x^(2)}+(y+1)at}{\sqrt{1-x^(2)}}}`

Now again consider ;

`{:\implies \quad \sf f(x,y,z)=e^(x+y)\cos (z)+(y+1)\sin^(-1)(x)at}`

Partial differentiating both sides w.r.t.y will yield ;

`{:\implies \quad \sf (\partial f)/(\partial y)=e^(x+y)\cos (z)+\sin^(-1)(x)at}`

Now , again consider ;

`{:\implies \quad \sf f(x,y,z)=e^(x+y)\cos (z)+(y+1)\sin^(-1)(x)at}`

Partial differentiating both sides w.r.t.z will yield ;

`{:\implies \quad \sf (\partial f)/(\partial z)=-e^(x+y)\sin (z)}`

This is the required answer

Used Concepts :-

• 
  `{\boxed{\bf{(d)/(dx)\{\sin^(-1)(x)\}=\frac{1}{\sqrt{1-x^(2)}}}}}`

• 
  `{\boxed{\bf{(d)/(dx)(e^x)=e^(x)}}}`

• 
  `{\boxed{\bf{(d)/(dx)\{\cos (x)\}=-\sin (x)}}}`