Question

Differentiate the following functions using first principal

$f(x) = {sin}^{ - 1}(2x + 3)$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {sin}^{ - 1}(2x + 3)$

Let assume that,

$\rm :\longmapsto\:f(x) = y = {sin}^{ - 1}(2x + 3)$

$\rm :\longmapsto\:y = {sin}^{ - 1}(2x + 3)$

$\rm :\longmapsto\:siny = 2x + 3$

$\rm :\longmapsto\:siny - 3= 2x$

$\bf\implies \:x = \dfrac{siny - 3}{2}$

So,

$\rm :\longmapsto\:x = g(y) = \dfrac{siny - 3}{2}$

So,

$\rm :\longmapsto\:g(y + h) = \dfrac{sin(y + h) - 3}{2}$

By using definition of First Principal, we have

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{g(y + h) - g(y)}{h}$

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{\dfrac{sin(y + h) - 3}{2} - \dfrac{siny - 3}{2} }{h}$

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{\dfrac{sin(y + h) - siny}{2}}{h}$

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{sin(y + h) - siny}{2h}$

We know,

$\red{ \boxed{ \sf{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, using this result, we get

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{2cos\bigg[\dfrac{y + h + y}{2} \bigg]sin\bigg[\dfrac{y + h - y}{2} \bigg]}{2h}$

$\rm :\longmapsto\:g'(y) = \displaystyle\lim_{h \to 0}\sf \frac{cos\bigg[\dfrac{2y + h}{2} \bigg]sin\bigg[\dfrac{h }{2} \bigg]}{h}$

$\rm :\longmapsto\:g'(y) =cosy \displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{h }{2} \bigg]}{\dfrac{h}{2} \times 2 }$

We know,

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1}}}$

So, using this, we get

$\rm :\longmapsto\:g'(y) =\dfrac{1}{2} cosy$

Now,

$\rm :\longmapsto\:f'(x) = \dfrac{1}{g'(y)}$

$\rm :\longmapsto\:f'(x) = \dfrac{2}{cosy}$

$\rm :\longmapsto\:f'(x) = \dfrac{2}{ \sqrt{1 - {sin}^{2}y } }$

$\rm :\longmapsto\:f'(x) = \dfrac{2}{ \sqrt{1 - {(2x + 3)}^{2}} }$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

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