Question
Differentiate the following from first principle:
$\frac{1}{\sqrt{3-\text{x}}}$
$\frac{1}{\sqrt{3-\text{x}}}$
$\text{f(x)}=\frac{1}{\sqrt{{3}-\text{x}}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{3-(\text{x+h)}}}-\frac{1}{\sqrt{3-\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}\sqrt{3-\text{(x+h)}\text{h}}}\times\frac{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}$
$\Big[ $Rationalising the numerator by $\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(3-\text{x})-\big(3-\text{(x+h)}\big)}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\bigg(\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\bigg)}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\big(\sqrt{3-\text{x}}+{\sqrt{3-\text{(x+h)}}}\big)}}$
$=\frac{1}{(3-\text{x})\times2\sqrt{3-\text{x}}}$
$=\frac{1}{2(3-\text{x})^\frac{3}{2}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If p is a real number and if the middle term in the expansion of $\Big(\frac{\text{p}}{2}+2\Big)^{8}$ is 1120, find p.
Find a, b and n in the expansion of
$(\text{a}+\text{b})^{\text{n}},$ if the first three terms in the expansion are 729, 7290 and 30375 respectively.