Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$

$\begin{aligned} & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot \frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right) \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{d}{d x}(3 x-5)^{\frac{1}{2}}-\frac{d}{d x}(3 x-5)^{-\frac{1}{2}}\right] \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2}(3 x-5)^{-\frac{1}{2}} \cdot \frac{d}{d x}(3 x-5)-\left(-\frac{1}{2}\right)(3 x-5)^{-\frac{3}{2}} \cdot \frac{d}{d x}(3 x-5)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2 \sqrt{3 x-5}} \cdot(3 \times 1-0)+\frac{1}{2(3 x-5)^{\frac{3}{2}}} \cdot(3 \times 1-0)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3}{2 \sqrt{3 x-5}}+\frac{3}{2(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15}{2}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3 x-5+1}{(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15(3 x-4)}{2(3 x-5)^{\frac{3}{2}}}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \text {. } \\ & \end{aligned}$