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

The answer to (\partial)/(\partial x)(4(y^2-x^2)ln(x+y)) is 4(-2xln(x+y)+y-x)

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

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

4 (- 2 x ln (x + y) + y - x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 4 \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)

= 4 (\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}

= 4 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2}))

Simplify

4 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2})) : 4 (- 2 x ln (x + y) + y - x)

= 4 (- 2 x ln (x + y) + y - x)

\int x^{2} \cdot 1 - x d x

\frac{1}{2 x} - \frac{8}{5 x ^{4}}

\int \frac{lo g _{e} ( x )}{x} d x

f (x) = ln (10 x)

e^{5 x + 5}

What is (\partial)/(\partial x)(4(y^2-x^2)ln(x+y)) ?
The answer to (\partial)/(\partial x)(4(y^2-x^2)ln(x+y)) is 4(-2xln(x+y)+y-x)