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Question
f\left(x,y\right)=e^{x+y}
Function
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\text{Find the first partial derivative with respect to }x
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\text{Find the first partial derivative with respect to }y
f_{x}=e^{x+y}
Simplify
f\left(x,y\right)=e^{x+y}
\text{Find the first partial derivative by treating the variable }y\text{ as a constant and differentiating with respect to }x
f_{x}=\frac{\partial}{\partial x}\left(e^{x+y}\right)
\text{Use the chain rule }\frac{\partial}{\partial x}(f(g))=\frac{\partial}{\partial g}(f(g))\times \frac{\partial}{\partial x}(g)\text{ where the }g\text{=}x+y\text{, to find the derivative}
f_{x}=\frac{\partial}{\partial g}\left(e^{g}\right)\times \frac{\partial}{\partial x}\left(x+y\right)
\text{Use }\frac{\partial}{\partial x} e^{x}=e^{x}\text{ to find derivative}
f_{x}=e^{g}\times \frac{\partial}{\partial x}\left(x+y\right)
Evaluate
More Steps
Evaluate
\frac{\partial}{\partial x}\left(x+y\right)
\text{Use differentiation rule }\frac{\partial}{\partial x}\left(f\left(x\right)\pm g\left(x\right)\right)=\frac{\partial}{\partial x}\left(f\left(x\right)\right)\pm \frac{\partial}{\partial x}(g(x))
\frac{\partial}{\partial x}\left(x\right)+\frac{\partial}{\partial x}\left(y\right)
\text{Use }\frac{\partial}{\partial x} x^{n}=n x^{n-1}\text{ to find derivative}
1+\frac{\partial}{\partial x}\left(y\right)
\text{Use }\frac{\partial}{\partial x}(c)=0\text{ to find derivative}
1+0
Removing 0 doesn't change the value,so remove it from the expression
1
f_{x}=e^{g}\times 1
Substitute back
f_{x}=e^{x+y}\times 1
Solution
f_{x}=e^{x+y}
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