Question

f(x,y)=x·e^{x^{2}y}
fx(x,y)=

Explanation:

Step1: Apply the product rule

The product rule states that if \(u(x)\) and \(v(x)\) are functions, then \((uv)^\prime=u^\prime v + uv^\prime\). Let \(u = x\) and \(v=e^{x^{2}y}\). Then \(u^\prime=\frac{\partial u}{\partial x}=1\).

Step2: Apply the chain rule for \(v = e^{x^{2}y}\)

The chain rule: if \(v = e^{g(x)}\), then \(v^\prime=e^{g(x)}\cdot g^\prime(x)\). Here \(g(x)=x^{2}y\), so \(g^\prime(x) = 2xy\). Then \(v^\prime=\frac{\partial v}{\partial x}=e^{x^{2}y}\cdot2xy\)

Step3: Combine using the product rule

\(f_x(x,y)=u^\prime v+uv^\prime=1\cdot e^{x^{2}y}+x\cdot e^{x^{2}y}\cdot2xy\)

Answer:

\(e^{x^{2}y}(1 + 2x^{2}y)\)