What is the derivative of-(40(9+5x^2))/((9-5x^2)^2) ?

The derivative of-(40(9+5x^2))/((9-5x^2)^2) is -(400x(5x^2+27))/((9-5x^2)^3)

What is the first derivative of-(40(9+5x^2))/((9-5x^2)^2) ?

The first derivative of-(40(9+5x^2))/((9-5x^2)^2) is -(400x(5x^2+27))/((9-5x^2)^3)

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derivative of

- \frac{40 ( 9 + 5 x ^{2} )}{( 9 - 5 x ^{2} ) ^{2}}

- \frac{400 x ( 5 x ^{2} + 27 )}{( 9 - 5 x ^{2} ) ^{3}}

Solution steps

\frac{d}{d x} (- \frac{40 ( 9 + 5 x ^{2} )}{( 9 - 5 x ^{2} ) ^{2}})

Take the constant out:

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

= - 40 \frac{d}{d x} (\frac{9 + 5 x ^{2}}{( 9 - 5 x ^{2} ) ^{2}})

Apply the Quotient Rule:

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

= - 40 \cdot \frac{\frac{d}{d x} ( 9 + 5 x ^{2} ) ( 9 - 5 x ^{2} ) ^{2} - \frac{d}{d x} ( ( 9 - 5 x ^{2} ) ^{2} ) ( 9 + 5 x ^{2} )}{( ( 9 - 5 x ^{2} ) ^{2} ) ^{2}}

\frac{d}{d x} (9 + 5 x^{2}) = 10 x

\frac{d}{d x} ((9 - 5 x^{2})^{2}) = - 20 x (9 - 5 x^{2})

= - 40 \cdot \frac{10 x ( 9 - 5 x ^{2} ) ^{2} - ( - 20 x ( 9 - 5 x ^{2} ) ) ( 9 + 5 x ^{2} )}{( ( 9 - 5 x ^{2} ) ^{2} ) ^{2}}

Simplify

- 40 \cdot \frac{10 x ( 9 - 5 x ^{2} ) ^{2} - ( - 20 x ( 9 - 5 x ^{2} ) ) ( 9 + 5 x ^{2} )}{( ( 9 - 5 x ^{2} ) ^{2} ) ^{2}} : - \frac{400 x ( 5 x ^{2} + 27 )}{( 9 - 5 x ^{2} ) ^{3}}

= - \frac{400 x ( 5 x ^{2} + 27 )}{( 9 - 5 x ^{2} ) ^{3}}

\int_{1}^{4} 2 x 5 + x^{2} d x

\frac{\partial}{\partial x} (e^{a x^{3}})

y = sec (5 x)

\int \frac{1}{sin ( θ ) ( cos ^{2} ( θ ) + 4 )} d x

\int e^{1.5 x} d x

What is the derivative of-(40(9+5x^2))/((9-5x^2)^2) ?
The derivative of-(40(9+5x^2))/((9-5x^2)^2) is -(400x(5x^2+27))/((9-5x^2)^3)
What is the first derivative of-(40(9+5x^2))/((9-5x^2)^2) ?
The first derivative of-(40(9+5x^2))/((9-5x^2)^2) is -(400x(5x^2+27))/((9-5x^2)^3)