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

The derivative of f(x)= 3/((x^2-5x)^2) is -(6(2x-5))/((x^2-5x)^3)

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

The first derivative of f(x)= 3/((x^2-5x)^2) is -(6(2x-5))/((x^2-5x)^3)

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

3 (- \frac{2}{( x ^{2} - 5 x ) ^{3}} (2 x - 5)) : - \frac{6 ( 2 x - 5 )}{( x ^{2} - 5 x ) ^{3}}

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

x \to 0 lim (\frac{x ^{\frac{2}{3}}}{x})

f (x) = tan^{2} (x), at x = \frac{π}{4}

\frac{sin ( x )}{3 x}

- lo g_{e} (1 - x)

y = 1 cos (2) (x)

What is the derivative of f(x)= 3/((x^2-5x)^2) ?
The derivative of f(x)= 3/((x^2-5x)^2) is -(6(2x-5))/((x^2-5x)^3)
What is the first derivative of f(x)= 3/((x^2-5x)^2) ?
The first derivative of f(x)= 3/((x^2-5x)^2) is -(6(2x-5))/((x^2-5x)^3)