Question

differentiate 3x+1÷x^3+4 with respect to x​

Answer 1

Answer:

okay

thanks for the points

Answer 2

$\tt{\frac{dy}{dx} = \frac{d}{dy} ( \frac{3x + 1}{x^{3}+4}) }$

Using quotient rule which says;

$\tt {\frac{d}{dy} (\frac{u}{v}) = \frac{ du. v - dv. u }{v^{2}} }$

Apply it in the question:

→ $\tt{\frac{d}{dy} (\frac{3x + 1}{x^{3}+4}) = \frac{d(3x+1).(x^{3}+4) - d(x^{3}+4)(3x+1)}{(x^{3}+4)^{2}}}$

→ $\tt{ \frac{3(x^{3} + 4) - 3x^{2}(3x+1)}{(x^{3}+4)^{2}}}$

→ $\tt{ \frac{12 - 3x² - 3x³}{(x^{3}+4)^{2}} }$