What is the derivative of ((6x-7)^2)/(x^4) ?

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

What is the first derivative of ((6x-7)^2)/(x^4) ?

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

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derivative of

\frac{( 6 x - 7 ) ^{2}}{x ^{4}}

\frac{4 ( 6 x - 7 ) ( - 3 x + 7 )}{x ^{5}}

Solution steps

\frac{d}{d x} (\frac{( 6 x - 7 ) ^{2}}{x ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( 6 x - 7 ) ^{2} ) x ^{4} - \frac{d}{d x} ( x ^{4} ) ( 6 x - 7 ) ^{2}}{( x ^{4} ) ^{2}}

\frac{d}{d x} ((6 x - 7)^{2}) = 12 (6 x - 7)

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

= \frac{12 ( 6 x - 7 ) x ^{4} - 4 x ^{3} ( 6 x - 7 ) ^{2}}{( x ^{4} ) ^{2}}

Simplify

\frac{12 ( 6 x - 7 ) x ^{4} - 4 x ^{3} ( 6 x - 7 ) ^{2}}{( x ^{4} ) ^{2}} : \frac{4 ( 6 x - 7 ) ( - 3 x + 7 )}{x ^{5}}

= \frac{4 ( 6 x - 7 ) ( - 3 x + 7 )}{x ^{5}}

\frac{d}{d x} (tan (x) + \frac{1}{x ^{2}})

\int 4 - x d x

x \to - \infty lim (x^{2} + 2)

\int_{1}^{2} 3 x^{2} d x

\int \frac{1}{x ^{2} + 4 x + 3} d x

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