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Differentiate f(x) = sin(3x+4) with respect to x, using first principal.

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Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = sin(3x + 4)$

So,

$\rm :\longmapsto\:f(x) = sin[3(x + h) + 4] = sin(3x + 3h + 4)$

By using definition of First Principal, we have

$\boxed{ \tt{ \: f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \: \: }}$

On substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{sin(3x + 3h + 4) - sin(3x + 4)}{h}$

We know,

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

So, on using this identity, we get

$\rm=\displaystyle\lim_{h \to 0} \sf \frac{2cos\bigg[\dfrac{3x + 3h + 4 + 3x + 4}{2} \bigg]sin\bigg[\dfrac{3x + 3h + 4 - 3x - 4}{2} \bigg]}{h}$

$\rm=\displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{6x + 6h + 8}{2} \bigg]sin\bigg[\dfrac{3h}{2} \bigg]}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0}cos\bigg[\dfrac{6x + 6h + 8}{2} \bigg]\displaystyle\lim_{h \to 0} \frac{2sin\bigg[\dfrac{3h}{2} \bigg]}{h}$

$\rm \: = \:cos\bigg[\dfrac{6x+ 8}{2} \bigg]\displaystyle\lim_{h \to 0} \frac{2sin\bigg[\dfrac{3h}{2} \bigg]}{ \dfrac{3h}{2} } \times \dfrac{3}{2}$

We know,

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

So, using this identity, we get

$\rm \: = \:cos(3x + 4) \times 3$

$\rm \: = \:3cos(3x + 4)$

Hence,

$\bf\implies \:\boxed{ \tt{ \: \dfrac{d}{dx}sin(3x + 4) = 3cos(3x + 4) \: \: }}$

Additional Information :-

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