What is (\partial)/(\partial y)(7(x+5y-4)^2) ?

The answer to (\partial)/(\partial y)(7(x+5y-4)^2) is 70(x+5y-4)

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(\partial)/(\partial y)(7(x+5y-4)^2)

\frac{\partial}{\partial y} (7 (x + 5 y - 4)^{2})

70 (x + 5 y - 4)

Solution steps

\frac{\partial}{\partial y} (7 (x + 5 y - 4)^{2})

Treat

x

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial y} ((x + 5 y - 4)^{2})

Apply the chain rule:

2 (x + 5 y - 4) \frac{\partial}{\partial y} (x + 5 y - 4)

= 2 (x + 5 y - 4) \frac{\partial}{\partial y} (x + 5 y - 4)

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

= 7 \cdot 2 (x + 5 y - 4) \cdot 5

Simplify

= 70 (x + 5 y - 4)

\int_{- 2}^{2} (4 x^{3} + 6 x^{2} - 8 x + 5) d x

\int 2400 x 3 + 2 x d x

\int_{0}^{1} \frac{2 x}{x ^{2} + 1} d x

\frac{s}{1 + s}

What is (\partial)/(\partial y)(7(x+5y-4)^2) ?
The answer to (\partial)/(\partial y)(7(x+5y-4)^2) is 70(x+5y-4)