Find the derivative of 1 + 6 x 3 5 + 4 x + 6 x 2 + 7 x 5 .

Differentiate using the Quotient Rule which states that&nbsp;ddx[f(x)g(x)]&nbsp;is&nbsp;g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2&nbsp;where&nbsp;f(x)=1+6x3&nbsp;and&nbsp;g(x)=5+4x+6x2+7x5.(5+4x+6x2+7x5)ddx[1+6x3]-(1+6x3)ddx[5+4x+6x2+7x5](5+4x+6x2+7x5)2Differentiate.18(5+4x+6x2+7x5)x2-(1+6x3)(4+12x+35x4)(5+4x+6x2+7x5)2Simplify.-84x7-x4-48x3-90x2+12x+4(7x5+6x2+4x+5)2

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College
Calculus and Analysis
Differential Calculus
Derivatives

kazue72949lard

Answered question

2022-05-13

Find the derivative of

\frac{1 + 6 x^{3}}{5 + 4 x + 6 x^{2} + 7 x^{5}}

Answer & Explanation

Oswaldo Rosales

Beginner2022-05-14Added 16 answers

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 1 + 6 x^{3}$ and $g (x) = 5 + 4 x + 6 x^{2} + 7 x^{5}$ .

$\frac{(5 + 4 x + 6 x^{2} + 7 x^{5}) \frac{d}{d x} [1 + 6 x^{3}] - (1 + 6 x^{3}) \frac{d}{d x} [5 + 4 x + 6 x^{2} + 7 x^{5}]}{{(5 + 4 x + 6 x^{2} + 7 x^{5})}^{2}}$

Differentiate.

$\frac{18 (5 + 4 x + 6 x^{2} + 7 x^{5}) x^{2} - (1 + 6 x^{3}) (4 + 12 x + 35 x^{4})}{{(5 + 4 x + 6 x^{2} + 7 x^{5})}^{2}}$

Simplify.

$- \frac{84 x^{7} - x^{4} - 48 x^{3} - 90 x^{2} + 12 x + 4}{{(7 x^{5} + 6 x^{2} + 4 x + 5)}^{2}}$

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