What is (\partial)/(\partial x)(-5xcos(x^2-5xy)) ?

The answer to (\partial)/(\partial x)(-5xcos(x^2-5xy)) is -5(cos(x^2-5xy)-xsin(x^2-5xy)(2x-5y))

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(\partial)/(\partial x)(-5xcos(x^2-5xy))

\frac{\partial}{\partial x} (- 5 x cos (x^{2} - 5 x y))

- 5 (cos (x^{2} - 5 x y) - x sin (x^{2} - 5 x y) (2 x - 5 y))

Solution steps

\frac{\partial}{\partial x} (- 5 x cos (x^{2} - 5 x y))

Treat

y

as a constant

Take the constant out:

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

= - 5 \frac{\partial}{\partial x} (x cos (x^{2} - 5 x y))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = cos (x^{2} - 5 x y)

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

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (cos (x^{2} - 5 x y)) = - sin (x^{2} - 5 x y) (2 x - 5 y)

= - 5 (1 \cdot cos (x^{2} - 5 x y) + (- sin (x^{2} - 5 x y) (2 x - 5 y)) x)

Simplify

= - 5 (cos (x^{2} - 5 x y) - x sin (x^{2} - 5 x y) (2 x - 5 y))

\frac{d ^{2}}{d x} (y) + 2 \frac{d y}{d x} + y = 0

\frac{\partial}{\partial x} (e^{- 2 z} y sin (x))

cos (θ), 0.5

153 x^{2} - 0.06 x^{3}

y = (\frac{3}{5} x)^{4}

What is (\partial)/(\partial x)(-5xcos(x^2-5xy)) ?
The answer to (\partial)/(\partial x)(-5xcos(x^2-5xy)) is -5(cos(x^2-5xy)-xsin(x^2-5xy)(2x-5y))