Question

Differentiate each of the following from first principles:
$\left(i\right)sin\sqrt{2}x$
$\left(ii\right)cos\sqrt{x}$
$\left(iii\right)tan\sqrt{x}$
$\left(iv\right)tan{x}^2$

Answer 1

Let $f\left(x\right)=sin\sqrt{x}$. Then $f(x+h)=sin\sqrt{(x+h)}$

$\frac{d\left(f\right(x\left)\right)}{dx}={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{sin\sqrt{2(x+h)}-sin\sqrt{2x}}{h}$

$={lim}_{h\to 0}\frac{2sin\left(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}\right)cos\left(\frac{\sqrt{2(x+h)+\sqrt{2x}}}{2}\right)}{h}$

$={lim}_{h\to 0}\frac{sin\left(\frac{\sqrt{2(x+h)}}{2}\right)\left(\sqrt{2(x+h)-\sqrt{2x}}\right)\left(\sqrt{2(x+h)+\sqrt{2x}}\right)}{\left(\sqrt{2(x+h)+\sqrt{2x}}\right)h}cos\left(\frac{\sqrt{2(x+h)+2x}}{2}\right)$

$={lim}_{h\to 0}\frac{sin\left(\frac{\sqrt{2(x+h)-\sqrt{2x}}}{2}\right)}{cos\left(\frac{\sqrt{2(x+h)-\sqrt{2x}}}{2}\right)}{lim}_{h\to 0}\frac{2(x+h)-2x}{\left(\sqrt{2(x+h)+\sqrt{2x}}\right)h}{lim}_{h\to 0}cos\left(\frac{\sqrt{2(x+h)+2x}}{2}\right)$

$=1\times \frac{2}{2\sqrt{2}x}cos\left(\sqrt{2x}\right)=\frac{cos\left(\sqrt{2x}\right)}{\sqrt{2x}}$

We have,
$f\left(x\right)=cos\sqrt{x}$

$\because f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{cos\sqrt{x+h}-cos\sqrt{x}}{h}$

$={lim}_{h\to 0}\frac{-2sin\left(\frac{\sqrt{(x+h)}-\sqrt{x}}{2}\right)(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)(\sqrt{x+h}+\sqrt{x})h}\times sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)$

Multiplying numerator and denominator by $\sqrt{(x+h)+\sqrt{x}}$

$={lim}_{h\to 0}\frac{-sin\left(\frac{\sqrt{(x+h)-\sqrt{x}}}{2}\right)}{\left(\frac{\sqrt{(x+h)}-\sqrt{x}}{2}\right)}\times \frac{x+h-x}{(\sqrt{x+h}+\sqrt{x})h}\times sin\left(\frac{\sqrt{(x+h)}+\sqrt{x}}{2}\right)$

$={lim}_{h\to 0}-1\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\times sin\left(\frac{\sqrt{(x+h)+\sqrt{x}}}{2}\right)$

$={lim}_{h\to 0}\frac{-1}{h(\sqrt{x+h}+\sqrt{x})}sin\frac{\sqrt{x+h}+\sqrt{x}}{2}$

$=\frac{-sin\sqrt{x}}{2\sqrt{x}}$

(iii) We have,

$f\left(x\right)=tan\sqrt{x}$

$\because f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$\because f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{tan\sqrt{(x+h)}-tan\sqrt{x}}{h}$

$={lim}_{h\to 0}\frac{sin\sqrt{x+h}-\sqrt{x}}{hcos\sqrt{x+h}cos\sqrt{x}}$

$={lim}_{h\to 0}\frac{sin\sqrt{x+h}-\sqrt{x}}{(x+h-x)cos\sqrt{x}.cos\sqrt{x+h}}={lim}_{h\to 0}\frac{sin(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})cos\sqrt{x}.cos\sqrt{x+h}}$

$={lim}_{h\to 0}\frac{sin(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x})}\times \frac{1}{(\sqrt{x+h}+\sqrt{x})cos\sqrt{x}.cos\sqrt{x+h}}$

$=1\times \frac{1}{2\sqrt{x}cos\sqrt{x}cos\sqrt{x+h}}$

$=\frac{1}{2\sqrt{x}co{s}^{2}x}=\frac{se{c}^{2}x}{2\sqrt{x}}$

(iv) We have,

$f\left(x\right)=tan{x}^2$

$\because f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{tan(x+h{)}^2-tan{x}^2}{h}={lim}_{h\to 0}\frac{\frac{sin(x+h{)}^2}{cos(x+h{)}^2-\frac{sin{x}^2}{cos{x}^2}}}{h}$

$={lim}_{h\to 0}\frac{sin(x+h{)}^{2}cos{x}^2-cos(x+h{)}^{2}sin{x}^2}{\frac{cos(x+h{)}^{2}cos{x}^2}{h}}={lim}_{h\to 0}\frac{sin\left(\right(x+h{)}^2-x^2)}{h.cos(x+h{)}^2.cos{x}^2}$

$={lim}_{h\to 0}\frac{sin(x^2+h^2+2hx-x^2)}{h.cos(x+h{)}^2.cos{x}^2}={lim}_{h\to 0}\frac{sin(h^2+2hx)}{h.cos(x+h{)}^2.cos{x}^2}$

$={lim}_{h\to 0}\frac{sinh}{h}\times \frac{(h+2x)}{cos(x+h{)}^2.cos{x}^2}$

$=1.\frac{2x}{co{s}^2(x{)}^2}$

$=2xse{c}^2{x}^2$