What is the derivative of y=((6-5x)/(x^2-1))^2 ?

The derivative of y=((6-5x)/(x^2-1))^2 is (2(6-5x)(5x^2-12x+5))/((x^2-1)^3)

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

The first derivative of y=((6-5x)/(x^2-1))^2 is (2(6-5x)(5x^2-12x+5))/((x^2-1)^3)

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derivative of

y = (\frac{6 - 5 x}{x ^{2} - 1})^{2}

\frac{2 ( 6 - 5 x ) ( 5 x ^{2} - 12 x + 5 )}{( x ^{2} - 1 ) ^{3}}

Solution steps

\frac{d}{d x} ((\frac{6 - 5 x}{x ^{2} - 1})^{2})

Apply the chain rule:

2 \cdot \frac{6 - 5 x}{x ^{2} - 1} \frac{d}{d x} (\frac{6 - 5 x}{x ^{2} - 1})

= 2 \cdot \frac{6 - 5 x}{x ^{2} - 1} \frac{d}{d x} (\frac{6 - 5 x}{x ^{2} - 1})

\frac{d}{d x} (\frac{6 - 5 x}{x ^{2} - 1}) = \frac{5 x ^{2} - 12 x + 5}{( x ^{2} - 1 ) ^{2}}

= 2 \cdot \frac{6 - 5 x}{x ^{2} - 1} \cdot \frac{5 x ^{2} - 12 x + 5}{( x ^{2} - 1 ) ^{2}}

Simplify

2 \cdot \frac{6 - 5 x}{x ^{2} - 1} \cdot \frac{5 x ^{2} - 12 x + 5}{( x ^{2} - 1 ) ^{2}} : \frac{2 ( 6 - 5 x ) ( 5 x ^{2} - 12 x + 5 )}{( x ^{2} - 1 ) ^{3}}

= \frac{2 ( 6 - 5 x ) ( 5 x ^{2} - 12 x + 5 )}{( x ^{2} - 1 ) ^{3}}

\int \frac{1}{x ( x - 7 )} d x

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

\int x sin (x) d x

\frac{d}{d x} (\frac{2 - sec ( x )}{tan ( x )})

y = 2 - 3 x, x = - 2, x = 2, y = 0

What is the derivative of y=((6-5x)/(x^2-1))^2 ?
The derivative of y=((6-5x)/(x^2-1))^2 is (2(6-5x)(5x^2-12x+5))/((x^2-1)^3)
What is the first derivative of y=((6-5x)/(x^2-1))^2 ?
The first derivative of y=((6-5x)/(x^2-1))^2 is (2(6-5x)(5x^2-12x+5))/((x^2-1)^3)