What is (\partial)/(\partial x)((tx(1-x))/(1-t^2)) ?

The answer to (\partial)/(\partial x)((tx(1-x))/(1-t^2)) is (t(1-2x))/(1-t^2)

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(\partial)/(\partial x)((tx(1-x))/(1-t^2))

\frac{\partial}{\partial x} (\frac{t x ( 1 - x )}{1 - t ^{2}})

\frac{t ( 1 - 2 x )}{1 - t ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{t x ( 1 - x )}{1 - t ^{2}})

Treat

t

as a constant

Take the constant out:

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

= \frac{t}{1 - t ^{2}} \frac{\partial}{\partial x} (x (1 - x))

Apply the Product Rule:

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

f = x, g = 1 - x

= \frac{t}{1 - t ^{2}} (\frac{\partial}{\partial x} (x) (1 - x) + \frac{\partial}{\partial x} (1 - x) x)

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

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

= \frac{t}{1 - t ^{2}} (1 \cdot (1 - x) + (- 1) x)

Simplify

\frac{t}{1 - t ^{2}} (1 \cdot (1 - x) + (- 1) x) : \frac{t ( 1 - 2 x )}{1 - t ^{2}}

= \frac{t ( 1 - 2 x )}{1 - t ^{2}}

\int (\frac{e ^{\frac{1}{x}}}{x ^{2}}) d x

\int \frac{( 10 x + 2 )}{( 3 x + 1 ) ( x - 1 )} d x

\int_{0}^{8} x^{2} 64 - x^{2} d x

y^{^{'}} - 5 y = 6 e^{7 t}

5 x - x^{2}

What is (\partial)/(\partial x)((tx(1-x))/(1-t^2)) ?
The answer to (\partial)/(\partial x)((tx(1-x))/(1-t^2)) is (t(1-2x))/(1-t^2)