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 ${\bf \nabla f}$ , 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 ;

• ${\boxed{\bf{\nabla F=\dfrac{\partial F}{\partial x}\hat{i}+\dfrac{\partial F}{\partial y}\hat{j}+\dfrac{\partial F}{\partial z}\hat{k}+\cdots}}}$

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 ;

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

Simplifying will yield ;

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

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 ;

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

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 ;

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

${:\implies \quad \bf \therefore \quad \underline{\underline{\nabla f=\bigg\{\dfrac{e^{x+y}\cos (z)\sqrt{1-x^{2}}+(y+1)at}{\sqrt{1-x^{2}}}\bigg\}\hat{i}+\bigg\{e^{x+y}\cos (z)+\sin^{-1}(x)at\bigg\}\hat{j}-\bigg\{e^{x+y}\sin (z)\bigg\}\hat{k}}}}$

This is the required answer

Used Concepts :-

• ${\boxed{\bf{\dfrac{d}{dx}\{\sin^{-1}(x)\}=\dfrac{1}{\sqrt{1-x^{2}}}}}}$

• ${\boxed{\bf{\dfrac{d}{dx}(e^x)=e^{x}}}}$

• ${\boxed{\bf{\dfrac{d}{dx}\{\cos (x)\}=-\sin (x)}}}$