What is (\partial)/(\partial x)((-3)/(y-3x)) ?

The answer to (\partial)/(\partial x)((-3)/(y-3x)) is -9/((y-3x)^2)

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(\partial)/(\partial x)((-3)/(y-3x))

\frac{\partial}{\partial x} (\frac{- 3}{y - 3 x})

- \frac{9}{( y - 3 x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{- 3}{y - 3 x})

Simplify

\frac{- 3}{y - 3 x} : - \frac{3}{y - 3 x}

= \frac{\partial}{\partial x} (- \frac{3}{y - 3 x})

Treat

y

as a constant

Take the constant out:

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

= - 3 \frac{\partial}{\partial x} (\frac{1}{y - 3 x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= - 3 \frac{\partial}{\partial x} ((y - 3 x)^{- 1})

Apply the chain rule:

- \frac{1}{( y - 3 x ) ^{2}} \frac{\partial}{\partial x} (y - 3 x)

= - \frac{1}{( y - 3 x ) ^{2}} \frac{\partial}{\partial x} (y - 3 x)

\frac{\partial}{\partial x} (y - 3 x) = - 3

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

Simplify

- 3 (- \frac{1}{( y - 3 x ) ^{2}} (- 3)) : - \frac{9}{( y - 3 x ) ^{2}}

= - \frac{9}{( y - 3 x ) ^{2}}

2 x^{2} + x - 1, at a = 0

\int_{0}^{3} x^{3} - x d x

f (x) = (4 x^{4} - 7 x^{2} + 6)

(x + \frac{1}{x})^{3}

x \to 2 lim (5 x^{3} - 9 x^{2} + 7)

What is (\partial)/(\partial x)((-3)/(y-3x)) ?
The answer to (\partial)/(\partial x)((-3)/(y-3x)) is -9/((y-3x)^2)