Question

Derivative of Tan (3x+4) by first principle

Answer 1

Answer:

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We need to know:

• $\displaystyle\lim_{\theta\rightarrow0}\frac{\tan\theta}{\theta} = 1$
• $\displaystyle\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}$

Then

$\displaystyle\frac{d}{dx}\tan(3x+4) = \lim_{h\rightarrow0}\frac1h\left[\tan(3(x+h)+4)-\tan(3x+4)\right] \\ \\=\lim_{h\rightarrow0}\frac1h\left[\frac{\tan(3x+4)+\tan(3h)}{1-\tan(3x+4)\tan(3h)}-\tan(3x+4)\right] \\ \\=\lim_{h\rightarrow0}\frac1h\left[\frac{\tan(3x+4)+\tan(3h)-\tan(3x+4)+\tan^2(3x+4)\tan(3h)}{1-\tan(3x+4)\tan(3h)}\right] \\ \\=\lim_{h\rightarrow0}\frac{\tan(3h)}h\left[\frac{1+\tan^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\right]$

$\displaystyle=\lim_{h\rightarrow0}\frac{\tan(3h)}h\left[\frac{1+\tan^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\right] \\ \\=\lim_{h\rightarrow0}3\times\frac{\tan(3h)}{3h}\times\frac{\sec^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\\ \\= 3\times 1\times \frac{\sec^2(3x+4)}{1-0}\\ \\=3\sec^2(3x+4)$

Answer 2

Derivative of Tan(3x+4) from first principle:

If f(x) is a function, then it's derivative from first principle can be given by

  $f'(x)= \lim_{h \to \0} \frac{f(x+h)-f(x)}{h} \\Here, f(x) = tan(3x+4) = \frac{sin(3x+4)}{cos(3x+4)} \\f(x+h)=tan(3x+3h+4)=\frac{sin(3x+3h+4)}{cos(3x+3h+4)} \\f'(x)= \lim_{h \to \ 0} \frac{\frac{sin(3x+3h+4)}{cos(3x+3h+4)}-\frac{sin(3x+4)}{cos(3x+4)}}{h} \\f'(x)= \lim_{h \to \ 0} \frac{sin(3x+3h+4)cos(3x+4)-cos(3x+3h+4)sin(3x+4)}{hcos(3x+3h+4)cos(3x+4)} \\f'(x)= \lim_{h \to \ 0} \frac{sin(3x+3h+4-3x-4)}{hcos(3x+3h+4)cos(3x+4)} \\sinAcosB-cosAsinB=sin(A-B)$

$f'(x)=3 \lim_{h \to \ 0} \frac{sin3h}{3h}  \lim_{h \to \ 0} \frac{1}{cos(3x+3h+4)cos(3x+4)} \\f'(x)=3\frac{1}{cos(3x+4)cos(3x+4)} \\f'(x)=3sec^{2} (3x+4)$

Therefore derivative of tan(3x+4) is 3sec$^{2}$(3x+4)