What is the derivative of y= 1/(4x^3)-6/(5x^4) ?

The derivative of y= 1/(4x^3)-6/(5x^4) is -3/(4x^4)+(24)/(5x^5)

What is the first derivative of y= 1/(4x^3)-6/(5x^4) ?

The first derivative of y= 1/(4x^3)-6/(5x^4) is -3/(4x^4)+(24)/(5x^5)

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derivative of

y = \frac{1}{4 x ^{3}} - \frac{6}{5 x ^{4}}

- \frac{3}{4 x ^{4}} + \frac{24}{5 x ^{5}}

Solution steps

\frac{d}{d x} (\frac{1}{4 x ^{3}} - \frac{6}{5 x ^{4}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{4 x ^{3}}) - \frac{d}{d x} (\frac{6}{5 x ^{4}})

\frac{d}{d x} (\frac{1}{4 x ^{3}}) = - \frac{3}{4 x ^{4}}

\frac{d}{d x} (\frac{6}{5 x ^{4}}) = - \frac{24}{5 x ^{5}}

= - \frac{3}{4 x ^{4}} - (- \frac{24}{5 x ^{5}})

Simplify

= - \frac{3}{4 x ^{4}} + \frac{24}{5 x ^{5}}

F (s) = ln (ln (3 s))

x^{- 2}

\int \frac{( x ^{4} )}{x ^{2} + 4} d x

\frac{d y}{d x} = \frac{- y ^{2}}{1 + y ^{2}} e^{- x}, y (0) = 1

x \to - 2 lim (h (x))

What is the derivative of y= 1/(4x^3)-6/(5x^4) ?
The derivative of y= 1/(4x^3)-6/(5x^4) is -3/(4x^4)+(24)/(5x^5)
What is the first derivative of y= 1/(4x^3)-6/(5x^4) ?
The first derivative of y= 1/(4x^3)-6/(5x^4) is -3/(4x^4)+(24)/(5x^5)