What is the derivative of y=(3(2x^5-7))/(x^4) ?

The derivative of y=(3(2x^5-7))/(x^4) is (6(x^5+14))/(x^5)

What is the first derivative of y=(3(2x^5-7))/(x^4) ?

The first derivative of y=(3(2x^5-7))/(x^4) is (6(x^5+14))/(x^5)

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derivative of

y = \frac{3 ( 2 x ^{5} - 7 )}{x ^{4}}

\frac{6 ( x ^{5} + 14 )}{x ^{5}}

Solution steps

\frac{d}{d x} (\frac{3 ( 2 x ^{5} - 7 )}{x ^{4}})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{2 x ^{5} - 7}{x ^{4}})

Apply the Quotient Rule:

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

= 3 \cdot \frac{\frac{d}{d x} ( 2 x ^{5} - 7 ) x ^{4} - \frac{d}{d x} ( x ^{4} ) ( 2 x ^{5} - 7 )}{( x ^{4} ) ^{2}}

\frac{d}{d x} (2 x^{5} - 7) = 10 x^{4}

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

= 3 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 7 )}{( x ^{4} ) ^{2}}

Simplify

3 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 7 )}{( x ^{4} ) ^{2}} : \frac{6 ( x ^{5} + 14 )}{x ^{5}}

= \frac{6 ( x ^{5} + 14 )}{x ^{5}}

f (x) = (x + 1)^{x}

\int_{0}^{2} 9 - x^{2} d x

g (x) = \frac{5 + 2 x}{7 - 8 x}

\int sin (k t) d t

- t^{2} + 9 t - 20

What is the derivative of y=(3(2x^5-7))/(x^4) ?
The derivative of y=(3(2x^5-7))/(x^4) is (6(x^5+14))/(x^5)
What is the first derivative of y=(3(2x^5-7))/(x^4) ?
The first derivative of y=(3(2x^5-7))/(x^4) is (6(x^5+14))/(x^5)