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

The answer to (\partial)/(\partial x)(5(y^2-x^2)ln(x+y)) is 5(-2xln(x+y)+y-x)

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

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

5 (- 2 x ln (x + y) + y - x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = y^{2} - x^{2}, g = ln (x + y)

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

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

\frac{\partial}{\partial x} (ln (x + y)) = \frac{1}{x + y}

= 5 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2}))

Simplify

5 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2})) : 5 (- 2 x ln (x + y) + y - x)

= 5 (- 2 x ln (x + y) + y - x)

t^{2} y^{^{''}} - 2 y = 0

\frac{d}{d x} (sin (x) cos (3 - x^{2}))

\frac{d}{d x} (e^{a x} a)

\frac{d y}{d x} = x^{3}

(x + y - 2) d x + (x - y + 4) d y = 0

What is (\partial)/(\partial x)(5(y^2-x^2)ln(x+y)) ?
The answer to (\partial)/(\partial x)(5(y^2-x^2)ln(x+y)) is 5(-2xln(x+y)+y-x)