What is (\partial)/(\partial y)(7e^{7e^{yx}+y}x) ?

The answer to (\partial)/(\partial y)(7e^{7e^{yx}+y}x) is 7xe^{7e^{yx}+y}(7e^{yx}x+1)

Popular >

(\partial)/(\partial y)(7e^{7e^{yx}+y}x)

\frac{\partial}{\partial y} (7 e^{7 e^{y x} + y} x)

7 x e^{7 e^{y x} + y} (7 e^{y x} x + 1)

Solution steps

\frac{\partial}{\partial y} (7 e^{7 e^{y x} + y} x)

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 x \frac{\partial}{\partial y} (e^{7 e^{y x} + y})

Apply the chain rule:

e^{7 e^{y x} + y} \frac{\partial}{\partial y} (7 e^{y x} + y)

= e^{7 e^{y x} + y} \frac{\partial}{\partial y} (7 e^{y x} + y)

\frac{\partial}{\partial y} (7 e^{y x} + y) = 7 e^{y x} x + 1

= 7 x e^{7 e^{y x} + y} (7 e^{y x} x + 1)

\frac{d}{d x} (- 4 sin (2 x) + 3 cos (3 x))

\int_{0}^{1} 6 e^{- 4 x} d x

\frac{d y}{d x} + \frac{2 x y}{1 + x ^{2}} = - 2 x y^{2}

\frac{28 s}{( s ^{2} + 11 )}

\frac{d y}{d x} + 4 tan (4 x) (y) = 3 cos (4 x)

What is (\partial)/(\partial y)(7e^{7e^{yx}+y}x) ?
The answer to (\partial)/(\partial y)(7e^{7e^{yx}+y}x) is 7xe^{7e^{yx}+y}(7e^{yx}x+1)