What is the derivative of y=((x^2+1)^5)/((1-x^2)^5) ?

The derivative of y=((x^2+1)^5)/((1-x^2)^5) is (20x(x^2+1)^4)/((1-x^2)^6)

What is the first derivative of y=((x^2+1)^5)/((1-x^2)^5) ?

The first derivative of y=((x^2+1)^5)/((1-x^2)^5) is (20x(x^2+1)^4)/((1-x^2)^6)

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derivative of

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

\frac{20 x ( x ^{2} + 1 ) ^{4}}{( 1 - x ^{2} ) ^{6}}

Solution steps

\frac{d}{d x} (\frac{( x ^{2} + 1 ) ^{5}}{( 1 - x ^{2} ) ^{5}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} ((x^{2} + 1)^{5}) = 10 x (x^{2} + 1)^{4}

\frac{d}{d x} ((1 - x^{2})^{5}) = - 10 x (1 - x^{2})^{4}

= \frac{10 x ( x ^{2} + 1 ) ^{4} ( 1 - x ^{2} ) ^{5} - ( - 10 x ( 1 - x ^{2} ) ^{4} ) ( x ^{2} + 1 ) ^{5}}{( ( 1 - x ^{2} ) ^{5} ) ^{2}}

Simplify

\frac{10 x ( x ^{2} + 1 ) ^{4} ( 1 - x ^{2} ) ^{5} - ( - 10 x ( 1 - x ^{2} ) ^{4} ) ( x ^{2} + 1 ) ^{5}}{( ( 1 - x ^{2} ) ^{5} ) ^{2}} : \frac{20 x ( x ^{2} + 1 ) ^{4}}{( 1 - x ^{2} ) ^{6}}

= \frac{20 x ( x ^{2} + 1 ) ^{4}}{( 1 - x ^{2} ) ^{6}}

f (x) = \frac{ln ( x )}{2} + 2

sin (x)

\int 30 d x

2 x^{2} - 10 x

n = 0 \sum \infty \frac{1}{ln ( n )}

What is the derivative of y=((x^2+1)^5)/((1-x^2)^5) ?
The derivative of y=((x^2+1)^5)/((1-x^2)^5) is (20x(x^2+1)^4)/((1-x^2)^6)
What is the first derivative of y=((x^2+1)^5)/((1-x^2)^5) ?
The first derivative of y=((x^2+1)^5)/((1-x^2)^5) is (20x(x^2+1)^4)/((1-x^2)^6)