- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey! 
I want to find the first and second derivative of the function $$\psi (\lambda )=f(\lambda x_1, \lambda x_2)$$ where $f(y_1, y_2)$ is twice differentiable and $(x_1, x_2)$ is arbitrary for fix.
I have done the following:
$$f(g(\lambda), h(\lambda)) : \\ \frac{\partial{f}}{\partial{\lambda}}=f_{y_1}\cdot g'(\lambda)+f_{y_2}\cdot h'(\lambda)=f_{y_1}\cdot x_1+f_{y_2}\cdot x_2$$
So, $\psi'(\lambda)=f_{y_1}\cdot x_1+f_{y_2}\cdot x_2$.
Is this correct? (Wondering)
Is the second derivative then the following:
\begin{align*}\frac{\partial^2{f}}{\partial{\lambda^2}}&=\left (f_{y_1y_1}\cdot g'(\lambda)+f_{y_1y_2}\cdot h'(\lambda)\right )\cdot x_1+\left (f_{y_2y_1}\cdot g'(\lambda)+f_{y_2y_2}\cdot h'(\lambda)\right )\cdot x_2 \\ & =\left (f_{y_1y_1}\cdot x_1+f_{y_1y_2}\cdot x_2\right )\cdot x_1+\left (f_{y_2y_1}\cdot x_1+f_{y_2y_2}\cdot x_2\right )\cdot x_2 \\ & =f_{y_1y_1}\cdot x_1^2+f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_1}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2 \\ & =f_{y_1y_1}\cdot x_1^2+2f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2\end{align*}
? (Wondering)
So, $\psi''(\lambda)=f_{y_1y_1}\cdot x_1^2+2f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2$.
Is this correct? (Wondering)
I want to find the first and second derivative of the function $$\psi (\lambda )=f(\lambda x_1, \lambda x_2)$$ where $f(y_1, y_2)$ is twice differentiable and $(x_1, x_2)$ is arbitrary for fix.
I have done the following:
$$f(g(\lambda), h(\lambda)) : \\ \frac{\partial{f}}{\partial{\lambda}}=f_{y_1}\cdot g'(\lambda)+f_{y_2}\cdot h'(\lambda)=f_{y_1}\cdot x_1+f_{y_2}\cdot x_2$$
So, $\psi'(\lambda)=f_{y_1}\cdot x_1+f_{y_2}\cdot x_2$.
Is this correct? (Wondering)
Is the second derivative then the following:
\begin{align*}\frac{\partial^2{f}}{\partial{\lambda^2}}&=\left (f_{y_1y_1}\cdot g'(\lambda)+f_{y_1y_2}\cdot h'(\lambda)\right )\cdot x_1+\left (f_{y_2y_1}\cdot g'(\lambda)+f_{y_2y_2}\cdot h'(\lambda)\right )\cdot x_2 \\ & =\left (f_{y_1y_1}\cdot x_1+f_{y_1y_2}\cdot x_2\right )\cdot x_1+\left (f_{y_2y_1}\cdot x_1+f_{y_2y_2}\cdot x_2\right )\cdot x_2 \\ & =f_{y_1y_1}\cdot x_1^2+f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_1}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2 \\ & =f_{y_1y_1}\cdot x_1^2+2f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2\end{align*}
? (Wondering)
So, $\psi''(\lambda)=f_{y_1y_1}\cdot x_1^2+2f_{y_1y_2}\cdot x_1\cdot x_2+f_{y_2y_2}\cdot x_2^2$.
Is this correct? (Wondering)