What is (\partial)/(\partial x)(y^{-2}(4y+9x^2)) ?

The answer to (\partial)/(\partial x)(y^{-2}(4y+9x^2)) is (18x)/(y^2)

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(\partial)/(\partial x)(y^{-2}(4y+9x^2))

\frac{\partial}{\partial x} (y^{- 2} (4 y + 9 x^{2}))

\frac{18 x}{y ^{2}}

Solution steps

\frac{\partial}{\partial x} (y^{- 2} (4 y + 9 x^{2}))

Treat

y

as a constant

Take the constant out:

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

= y^{- 2} \frac{\partial}{\partial x} (4 y + 9 x^{2})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= y^{- 2} (\frac{\partial}{\partial x} (4 y) + \frac{\partial}{\partial x} (9 x^{2}))

\frac{\partial}{\partial x} (4 y) = 0

\frac{\partial}{\partial x} (9 x^{2}) = 18 x

= y^{- 2} (0 + 18 x)

Simplify

y^{- 2} (0 + 18 x) : \frac{18 x}{y ^{2}}

= \frac{18 x}{y ^{2}}

f (x) = (x^{3} + x)^{6}

\frac{\partial}{\partial x} (sin (3 t))

f (x) = cos (x^{4})

\frac{\partial}{\partial s} (\frac{( r + s )}{t})

y = 7 ln (\frac{e ^{x} + e ^{- x}}{2}), (0, 0)

What is (\partial)/(\partial x)(y^{-2}(4y+9x^2)) ?
The answer to (\partial)/(\partial x)(y^{-2}(4y+9x^2)) is (18x)/(y^2)