$\frac{3 x+4}{5 x^{2}-7 x+9}$


$\frac{3 x+4}{5 x^{2}-7 x+9}$


Given $\mathrm{y}=\frac{3 \mathrm{x}+4}{5 \mathrm{x}^{2}-7 \mathrm{x}+9}$

Applying quotient rule of differentiation that is

$\Rightarrow \frac{d}{d x}\left(\frac{t}{y}\right)=\frac{y \cdot \frac{d t}{d x}-t \cdot \frac{d y}{d x}}{y^{2}}$

Applying the rule

$\Rightarrow \frac{d y}{d x}=\frac{\left(5 x^{2}-7 x+9\right) \frac{d}{d x}(3 x+4)-(3 x+4) \frac{d}{d x}\left(5 x^{2}-7 x+9\right)}{\left(5 x^{2}-7 x+9\right)^{2}}$

On differentiation we get

$=\left(15 x^{2}-21 x+27-30 x^{2}+21 x-40 x+28\right) /\left(5 x^{2}-7 x+9\right)^{2}$

$=\left(-15 x^{2}-40 x+55\right) /\left(5 x^{2}-7 x+9\right)^{2}$

$=\left(55-40 x-15 x^{2}\right) /\left(5 x^{2}-7 x+9\right)^{2}$

Hence, the required answer is

$\left(55-40 x-15 x^{2}\right) /\left(5 x^{2}-7 x+9\right)^{2}$

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