What is the value of $g'(1)$? Suppose $f(x,y)$ is real valued function for which $f(1,1)=1$ and its gradient at this point point is given by $\nabla f(1,1)=(-4,5)$.
Define a function $g(t)$ by $g(t)=f(t,f(t^2,t^3))$. Then what is the derivative of $g$ at $t=1$?
I deduce that $g(1)=f(1,1)=1$.
How to find $g'(1)$ ? Any help or hint.
Thanks in advance.
 A: You can use total differentiation:
\begin{align*}
g'(t) = \frac{\partial f}{\partial x}\frac{dt}{dt} + \frac{\partial f}{\partial y} \left(\frac{\partial f}{\partial x}\frac{dt^2}{dt}+\frac{\partial f}{\partial y}\frac{dt^3}{dt}\right).
\end{align*}
Note that $\frac{\partial f}{\partial x}(1,1) = -4$, $\frac{\partial f}{\partial y}(1,1) = 5$, and $x=t,y=f(t^2,t^3)$, so for $t=1$ you have $x=1$ and $y=f(1,1) = 1$. Thus
$$g'(1) = -4\cdot 1+5(-4\cdot 2\cdot 1+5\cdot 3\cdot 1) = 35-4=31.$$
A: $$dg(t)
\\=f_x(t,f(t^2,t^3))\,dt+f_y(t,f(t^2,t^3))\,df(t^2,t^3)
\\=f_x(t,f(t^2,t^3))\,dt+f_y(t,f(t^2,t^3))(2tf_x(t^2,t^3)+3t^2f_y(t^2,t^3))\,dt.$$
The rest is yours.

More legibly,
$$\frac{dg}{dt}=f_x+f_y\,\frac{df}{dt}=f_x+f_y(2tf_x+3t^2f_y).$$
A: Let $$x=t\\
y=t^2\\
z=t^3$$
Therefore,
$$g(t)=f(x,f(y,z))\\
\implies g'(t)=\dfrac{\partial f(x,f(y,z))}{\partial x}\dfrac{dx}{dt}+\underbrace{\dfrac{\partial f(x,f(y,z))}{\partial f(y,z)}}_{\begin{aligned}{\text{Partial differentiation with}}\\{\text{respect to second coordinate}}\end{aligned}}\left(\dfrac{\partial f(y,z)}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial f(y,z)}{\partial z}\dfrac{dz}{dt}\right)$$
Putting $t=1$ we get,
\begin{align*}
g'(t)&=\Big(-4\Big)\Big(1\Big)+\Big(5\Big)\Big((-4)(2)+(5)(3)\Big)\\
&=-4+5\times7\\
&=\boxed{31}
\end{align*}
A: According to the defination of gradient 
$$\nabla{f(x, y)} = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} f_x & f_y \end{bmatrix}$$
It is given that $$\nabla f(1,1) = \begin{bmatrix} -4 & 5 \end{bmatrix}$$
Hence $f_x = -4 \text{ and } f_y = 5$ at $x=1$ and $y=1$ respectively.
Now get back to the original question: [Assume: $a = t, b= t^2, c=t^3$]
$$\nabla f(a, f(b, c)) = \begin{bmatrix} \frac{\partial f(a, f(b, c))}{\partial x} & \frac{\partial f(a, f(b, c))}{\partial y} \end{bmatrix}$$
$$\nabla f(a, f(b, c)) = \begin{bmatrix} \frac{\partial f(a, f(b, c))}{\partial a} . \frac{\partial a}{\partial x} & \frac{\partial f(a, f(b, c))}{\partial f(b, c)} . \frac{\partial f(b, c)}{\partial y} \end{bmatrix}$$
$$\nabla f(a, f(b, c)) = \begin{bmatrix} \frac{\partial f(a, f(b, c))}{\partial a} . \frac{\partial a}{\partial x} & \frac{\partial f(a, f(b, c))}{\partial f(b, c)} . (\frac{\partial f(b, c)}{\partial b} . \frac{\partial b}{\partial y} + \frac{\partial f(b, c)}{\partial c} . \frac{\partial c}{\partial y}) \end{bmatrix}$$
Substitute the value of $a, b$ and $c$.
$$\nabla f(t, f(t^2, t^3)) = \begin{bmatrix} \frac{\partial f(t, f(t^2, t^3))}{\partial t} . \frac{\partial t}{\partial x} & \frac{\partial f(t, f(t^2, t^3))}{\partial f(t^2, t^3)} . (\frac{\partial f(t^2, t^3)}{\partial t^2} . \frac{\partial t^2}{\partial y} + \frac{\partial f(t^2, t^3)}{\partial t^3} . \frac{\partial t^3}{\partial y}) \end{bmatrix}$$
Substitute of value of t
$$\nabla f(t, f(t^2, t^3))|_{\ t=1} = \begin{bmatrix} 1.f_x & f_y .(f_x . 2 + f_y. 3) \end{bmatrix}$$
where . represents the scalar multiplication.
Substitute the value fo partial derivatives we get
$$\nabla f(t, f(t^2, t^3))|_{\ t=1} = \begin{bmatrix} (1)(-4) & (5)((-4)(2) + (5)(3)) \end{bmatrix}$$
$$\nabla f(t, f(t^2, t^3))|_{\ t=1} = \begin{bmatrix} -4 & 35\end{bmatrix}$$
