What is (\partial)/(\partial x)(6e^{x^5y^4}) ?

The answer to (\partial)/(\partial x)(6e^{x^5y^4}) is 30e^{x^5y^4}x^4y^4

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(\partial)/(\partial x)(6e^{x^5y^4})

\frac{\partial}{\partial x} (6 e^{x^{5} y^{4}})

30 e^{x^{5} y^{4}} x^{4} y^{4}

Solution steps

\frac{\partial}{\partial x} (6 e^{x^{5} y^{4}})

Treat

y

as a constant

Take the constant out:

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

= 6 \frac{\partial}{\partial x} (e^{x^{5} y^{4}})

Apply the chain rule:

e^{x^{5} y^{4}} \frac{\partial}{\partial x} (x^{5} y^{4})

= e^{x^{5} y^{4}} \frac{\partial}{\partial x} (x^{5} y^{4})

\frac{\partial}{\partial x} (x^{5} y^{4}) = 5 y^{4} x^{4}

= 6 e^{x^{5} y^{4}} \cdot 5 y^{4} x^{4}

Simplify

= 30 e^{x^{5} y^{4}} x^{4} y^{4}

x \to 0 lim (\frac{x}{tan ( 7 x )})

\frac{\partial}{\partial x} (e^{- x yz})

\frac{\partial}{\partial x} (ln (x + 3))

3 (1 + x) \frac{d y}{d x} = 2 x y (y^{3} + 1)

f (x) = \frac{8}{x} - \frac{4}{x}

What is (\partial)/(\partial x)(6e^{x^5y^4}) ?
The answer to (\partial)/(\partial x)(6e^{x^5y^4}) is 30e^{x^5y^4}x^4y^4