What is (\partial)/(\partial x)(7xsin(5x^2y)) ?

The answer to (\partial)/(\partial x)(7xsin(5x^2y)) is 7(sin(5x^2y)+10x^2ycos(5x^2y))

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(\partial)/(\partial x)(7xsin(5x^2y))

\frac{\partial}{\partial x} (7 x sin (5 x^{2} y))

7 (sin (5 x^{2} y) + 10 x^{2} y cos (5 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (7 x sin (5 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial x} (x sin (5 x^{2} y))

Apply the Product Rule:

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

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

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

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

\frac{\partial}{\partial x} (sin (5 x^{2} y)) = cos (5 x^{2} y) \cdot 10 y x

= 7 (1 \cdot sin (5 x^{2} y) + cos (5 x^{2} y) \cdot 10 y xx)

Simplify

7 (1 \cdot sin (5 x^{2} y) + cos (5 x^{2} y) \cdot 10 y xx) : 7 (sin (5 x^{2} y) + 10 x^{2} y cos (5 x^{2} y))

= 7 (sin (5 x^{2} y) + 10 x^{2} y cos (5 x^{2} y))

\int x x x + 1 d x

y^{^{''}} + 4 y^{^{'}} + 3 y = 15 t e^{3 t}

x^{2} = 4 y, 3 x - 4 y + 4 = 0

y^{^{'}} + y = e^{- t} sin (t), y (0) = 1

y^{^{'}} = y (1 - y)

What is (\partial)/(\partial x)(7xsin(5x^2y)) ?
The answer to (\partial)/(\partial x)(7xsin(5x^2y)) is 7(sin(5x^2y)+10x^2ycos(5x^2y))