Continuity and Differentiability — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability2 Marks
Question
Differentiate w.r.t. x, the function,$\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^{2}+4}}$
✓
Answer
Let y = $\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^{2}+4}}$ or y = $(3x+2)^\frac12+(2x^2+4)^{-\frac12}$ Note that this function is defined at all real numbers x > -$\frac{2}{3}$. Therefore $\frac{d y}{d x}=\frac{1}{2}(3 x+2)^{\frac{1}{2}-1}$.$\frac{d}{d x}(3 x+2)+\left(-\frac{1}{2}\right)\left(2 x^{2}+4\right)^{-\frac{1}{2}-1} \cdot \frac{d}{d x}\left(2 x^{2}+4\right)$ = $\frac{1}{2}(3 x+2)^{-\frac{1}{2}} \cdot(3)-\left(\frac{1}{2}\right)\left(2 x^{2}+4\right)^{-\frac{3}{2}} \cdot 4 x$ = $\frac{3}{2 \sqrt{3 x+2}}-\frac{2 x}{\left(2 x^{2}+4\right)^{\frac{3}{2}}}$ This is defined for all real numbers x > -$\frac{2}{3}$.
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