What is the derivative of #sqrt((7x+2) / (6x−3)) #?

Question

Aurora Alvarado · Accepted Answer

#=-\frac{11\sqrt{6x-3}}{6(3x-1)^2\sqrt{7x+2}}#

We can rewrite the function as so #(\frac{7x+2}{6x-3})^{1/2}#

We can then use the chain rule #1/2(\frac{7x+2}{6x-3})^{-1/2}=\frac{1}{2(\frac{7x+2}{6x-3})^{1/2}#

We then have to multiply this by the function inside the radical. To do this, we have to use the quotient rule (low dee high minus high dee low down below the square must go).

#\frac{(6x-3)(7)-(7x+2)(6)}{(6x-3)^2}# #=\frac{42x-21-42x-12}{(6x-3)^2}# #=-\frac{33}{(6x-3)^2}# #=-\frac{33}{9(2x-1}^2}# #=-\frac{11}{3(2x-1)^2#

Multiplying the two derivatives we got, we get #\frac{1}{2(\frac{7x+2}{6x-3})^{1/2}}(-\frac{11}{3(2x-1)^2))# #=-\frac{(6x-3)^{1/2}}{2(7x+2)^{1/2}}\cdot\frac{11}{3(2x-1)^2}# #=-\frac{11(6x-3)^{1/2}}{6(3x-1)^2(7x+2)^{1/2}}#

Adam Adkins · Answer

undefined To find the derivative of $\sqrt{\frac{7x+2}{6x-3}}$, you can use the chain rule and the quotient rule. The derivative is:

$$ \frac{d}{dx} \sqrt{\frac{7x+2}{6x-3}} = \frac{1}{2\sqrt{\frac{7x+2}{6x-3}}} \cdot \frac{d}{dx} \left(\frac{7x+2}{6x-3}\right) $$

$$ = \frac{1}{2\sqrt{\frac{7x+2}{6x-3}}} \cdot \left(\frac{(6x-3) \cdot \frac{d}{dx}(7x+2) - (7x+2) \cdot \frac{d}{dx}(6x-3)}{(6x-3)^2}\right) $$

$$ = \frac{1}{2\sqrt{\frac{7x+2}{6x-3}}} \cdot \left(\frac{(6x-3) \cdot 7 - (7x+2) \cdot 6}{(6x-3)^2}\right) $$

$$ = \frac{1}{2\sqrt{\frac{7x+2}{6x-3}}} \cdot \left(\frac{42x - 21 - 42x - 12}{(6x-3)^2}\right) $$

$$ = \frac{1}{2\sqrt{\frac{7x+2}{6x-3}}} \cdot \left(\frac{-33}{(6x-3)^2}\right) $$

$$ = \frac{-33}{2(6x-3)^2 \sqrt{\frac{7x+2}{6x-3}}} $$