Question
Function
Find the first partial derivative with respect to x
Find the first partial derivative with respect to y
∂x∂w=xxex+xln(y)+y
Simplify
w=ex+xln(y)+yln(x)
Find the first partial derivative by treating the variable y as a constant and differentiating with respect to x
∂x∂w=∂x∂(ex+xln(y)+yln(x))
Use differentiation rule ∂x∂(f(x)±g(x))=∂x∂(f(x))±∂x∂(g(x))
∂x∂w=∂x∂(ex)+∂x∂(xln(y))+∂x∂(yln(x))
Use ∂x∂ex=ex to find derivative
∂x∂w=ex+∂x∂(xln(y))+∂x∂(yln(x))
Evaluate
More Steps
Evaluate
∂x∂(xln(y))
Use differentiation rule ∂x∂(f(x)×g(x))=∂x∂(f(x))×g(x)+f(x)×∂x∂(g(x))
∂x∂(x)×ln(y)+x×∂x∂(ln(y))
Use ∂x∂xn=nxn−1 to find derivative
1×ln(y)+x×∂x∂(ln(y))
Evaluate
ln(y)+x×∂x∂(ln(y))
Use ∂x∂(c)=0 to find derivative
ln(y)+x×0
Evaluate
ln(y)+0
Removing 0 doesn't change the value,so remove it from the expression
ln(y)
∂x∂w=ex+ln(y)+∂x∂(yln(x))
Evaluate
More Steps
Evaluate
∂x∂(yln(x))
Use differentiation rule ∂x∂(f(x)×g(x))=∂x∂(f(x))×g(x)+f(x)×∂x∂(g(x))
∂x∂(y)×ln(x)+y×∂x∂(ln(x))
Use ∂x∂(c)=0 to find derivative
0×ln(x)+y×∂x∂(ln(x))
Evaluate
0+y×∂x∂(ln(x))
Evaluate
0+y×x1
Evaluate
0+xy
Removing 0 doesn't change the value,so remove it from the expression
xy
∂x∂w=ex+ln(y)+xy
Reduce fractions to a common denominator
∂x∂w=xexx+xln(y)×x+xy
Calculate
∂x∂w=xxex+xxln(y)+xy
Solution
∂x∂w=xxex+xln(y)+y
Show Solution
Solve the equation
w=ex+ln(yxxy)
Evaluate
w=ex+xln(y)+yln(x)
Solution
w=ex+ln(yxxy)
Show Solution