Question

If $y={\log }_{x^2+4}(7x^2-5x+1),$ then $\frac{dy}{dx}=$

• A. $\frac{1}{{\log }_e(x^2+4)}(\frac{14x-5}{7x^2-5x+1}-\frac{2xy}{x^2+4})$
• B. $\frac{1}{{\log }_e(x^2+4)}(\frac{14x-5}{7x^2-5x+1}+\frac{2xy}{x^2+4})$
• C. $-\frac{1}{{\log }_e(x^2+4)}(\frac{14x-5}{7x^2-5x+1}-\frac{2xy}{x^2+4})$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{{\log }_e(x^2+4)}(\frac{14x-5}{7x^2-5x+1}-\frac{2xy}{x^2+4})$

$y={\log }_{x^2+4}(7x^2-5x+1)=\frac{{\log }_e(7x^2-5x+1)}{{\log }_e(x^2+4)}=\frac{{\log }_{e}f\left(x\right)}{{\log }_{e}g\left(x\right)}$

$\therefore \frac{dy}{dx}=\frac{{\log }_{e}g\left(x\right)\frac{f^{\prime }\left(x\right)}{f\left(x\right)}-{\log }_{e}f\left(x\right)\frac{g^{\prime }\left(x\right)}{g\left(x\right)}}{{\left[{\log }_{e}g\left(x\right)\right]}^2}$

$=\frac{1}{{\log }_{e}g\left(x\right)}[\frac{f^{\prime }\left(x\right)}{f\left(x\right)}-y\frac{g^{\prime }\left(x\right)}{g\left(x\right)}]$

$=\frac{1}{{\log }_e(x^2+4)}(\frac{14x-5}{7x^2-5x+1}-\frac{2xy}{x^2+4})$