What is the derivative of y=(2(2x^5-9))/(x^4) ?

The derivative of y=(2(2x^5-9))/(x^4) is (4(x^5+18))/(x^5)

What is the first derivative of y=(2(2x^5-9))/(x^4) ?

The first derivative of y=(2(2x^5-9))/(x^4) is (4(x^5+18))/(x^5)

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derivative of

y = \frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

\frac{4 ( x ^{5} + 18 )}{x ^{5}}

Solution steps

\frac{d}{d x} (\frac{2 ( 2 x ^{5} - 9 )}{x ^{4}})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{2 x ^{5} - 9}{x ^{4}})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d}{d x} ( 2 x ^{5} - 9 ) x ^{4} - \frac{d}{d x} ( x ^{4} ) ( 2 x ^{5} - 9 )}{( x ^{4} ) ^{2}}

\frac{d}{d x} (2 x^{5} - 9) = 10 x^{4}

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

= 2 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 9 )}{( x ^{4} ) ^{2}}

Simplify

2 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 9 )}{( x ^{4} ) ^{2}} : \frac{4 ( x ^{5} + 18 )}{x ^{5}}

= \frac{4 ( x ^{5} + 18 )}{x ^{5}}

\int csc^{3} (x) sec (x) d x

\frac{d y}{d t} = y^{3}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 cos (3 t) - 20 sin (3 t)

f (x) = - x^{2} + 25, at x = 3

\int (1 + 2 x)^{- 1} d x

What is the derivative of y=(2(2x^5-9))/(x^4) ?
The derivative of y=(2(2x^5-9))/(x^4) is (4(x^5+18))/(x^5)
What is the first derivative of y=(2(2x^5-9))/(x^4) ?
The first derivative of y=(2(2x^5-9))/(x^4) is (4(x^5+18))/(x^5)