Derivative of tan2x by First Principle

The derivative of tan2x is equal to 2sec²2x. Here we will prove this by the first principle. In this post, we will learn how to differentiate tan2x with respect to x.

The derivative of tan2x formula is given by $\dfrac{d}{dx}(\tan 2x)$ = 2sec²2x.

Differentiate tan2x Using First Principle

To differentiate tan2x using the first principle, let us recall its definition first. The differentiation of f(x) by first principle is equal to the limit

$\dfrac{d}{dx}\big(f(x) \big)$ = lim_h→0 $\dfrac{f(x+h)-f(x)}{h}$.

In the above formula, we put f(x)=tan2x. Therefore,

$\dfrac{d}{dx}\big(\tan 2x \big)$ = lim_h→0 $\dfrac{\tan 2(x+h)- \tan 2x}{h}$

= lim_h→0 $\dfrac{\tan (2x+2h)- \tan 2x}{h}$

Now using the trigonometric formula tanx = $\dfrac{\sin x}{\cos x}$, we get that

$\dfrac{d}{dx}\big(\tan 2x \big)$ = lim_h→0 $\dfrac{\frac{\sin (2x+2h)}{\cos (2x+2h)}- \frac{\sin 2x}{\cos 2x}}{h}$

= lim_h→0 $\frac{\sin(2x+2h)\cos 2x -\sin 2x\cos(2x+2h)}{h \cos (2x+2h)\cos 2x}$

= lim_h→0 $\dfrac{\sin(2x+2h-2x)}{h\cos (2x+2h)\cos 2x}$ using the formula sin(a-b) = sina cosb -cosa sinb

= lim_h→0 $\dfrac{\sin 2h}{h\cos (2x+2h)\cos 2x}$

= 2 lim_h→0 $\dfrac{\sin 2h}{2h}$ × lim_h→0 $\dfrac{1}{\cos (2x+2h)\cos 2x}$

[Let 2h=t. Then t→0 as h→0.]

= 2 lim_t→0 $\dfrac{\sin t}{t}$ × lim_h→0 $\dfrac{1}{\cos (2x+2h)\cos 2x}$

= 2 × 1 × $\dfrac{1}{\cos (2x+0)\cos 2x}$ by the formula lim_x→0 (sinx)/x =1.

= $\dfrac{2}{\cos^2 2x}$

= 2sec² 2x as the reciprocal of cosx is secx.

Therefore, the derivative of tan2x is 2sec²2x, and this is proved by the first principle of derivatives.

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