What is the derivative of h(x)=((3x-5))/(1-x^2) ?

The derivative of h(x)=((3x-5))/(1-x^2) is (3+3x^2-10x)/((1-x^2)^2)

What is the first derivative of h(x)=((3x-5))/(1-x^2) ?

The first derivative of h(x)=((3x-5))/(1-x^2) is (3+3x^2-10x)/((1-x^2)^2)

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derivative of

h (x) = \frac{( 3 x - 5 )}{1 - x ^{2}}

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

Solution steps

\frac{d}{d x} (\frac{3 x - 5}{1 - x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x - 5 ) ( 1 - x ^{2} ) - \frac{d}{d x} ( 1 - x ^{2} ) ( 3 x - 5 )}{( 1 - x ^{2} ) ^{2}}

\frac{d}{d x} (3 x - 5) = 3

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

= \frac{3 ( 1 - x ^{2} ) - ( - 2 x ) ( 3 x - 5 )}{( 1 - x ^{2} ) ^{2}}

Simplify

\frac{3 ( 1 - x ^{2} ) - ( - 2 x ) ( 3 x - 5 )}{( 1 - x ^{2} ) ^{2}} : \frac{3 + 3 x ^{2} - 10 x}{( 1 - x ^{2} ) ^{2}}

= \frac{3 + 3 x ^{2} - 10 x}{( 1 - x ^{2} ) ^{2}}

\frac{d}{d x} (3 x^{2})

\frac{d}{d x} (2 x^{3} + \frac{5 x + 1}{3 x - 5} - \frac{2}{x ^{3}})

y^{^{'}} + y = 8 x e^{x}

\frac{d}{d x} (\frac{1}{12} (x + 2)^{3} + (x + 2)^{- 1})

x \to 0 lim (\frac{7 ^{x} - 9 ^{x}}{x})

What is the derivative of h(x)=((3x-5))/(1-x^2) ?
The derivative of h(x)=((3x-5))/(1-x^2) is (3+3x^2-10x)/((1-x^2)^2)
What is the first derivative of h(x)=((3x-5))/(1-x^2) ?
The first derivative of h(x)=((3x-5))/(1-x^2) is (3+3x^2-10x)/((1-x^2)^2)