Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\frac{\left(x^3-5\right)^5}{\left(x^3+3\right)^3}\right] \\ & =\frac{\left(x^3+3\right)^3 \cdot \frac{d}{d x}\left(x^3-5\right)^5-\left(x^3-5\right)^5 \cdot \frac{d}{d x}\left(x^3+3\right)^3}{\left[\left(x^3+3\right)^3\right]^2} \\ & =\frac{\left(x^3+3\right)^3 \times 5\left(x^3-5\right)^4 \cdot \frac{d}{d x}\left(x^3-5\right)-\left(x^3-5\right)^5 \times 3\left(x^3+3\right)^2 \cdot \frac{d}{d x}\left(x^3+3\right)}{\left(x^3+3\right)^6} \\ & =\frac{5\left(x^3+3\right)^3\left(x^3-5\right)^4 \cdot\left(3 x^2-0\right)-3\left(x^3-5\right)^5\left(x^3+3\right)^2 \cdot\left(3 x^2+0\right)}{\left(x^3+3\right)^6}\end{aligned}$

$=\frac{3 x^2\left(x^3+3\right)^2\left(x^3-5\right)^4\left[5\left(x^3+3\right)-3\left(x^3-5\right)\right]}{\left(x^3+3\right)^6}$

$\begin{aligned} & =\frac{3 x^2\left(x^3-5\right)^4\left(5 x^3+15-3 x^3+15\right)}{\left(x^3+3\right)^4} \\ & =\frac{3 x^2\left(x^3-5\right)^4\left(2 x^3+30\right)}{\left(x^3+3\right)^4} \\ & =\frac{6 x^2\left(x^3+15\right)\left(x^3-5\right)^4}{\left(x^3+3\right)^4}\end{aligned}$