Derivative of tan(x) - Proof and Explanation

Proof #1

Explanation

1. The proof starts with the definition of the tangent function:

   $\tan x=\frac{\sin (x)}{\cos (x)}$

2. It then uses the previously proven derivatives of sine and cosine:

   $\frac{d}{dx}\sin (x)=\cos (x)$ $\frac{d}{dx}\cos (x)=-\sin (x)$

3. The quotient rule for derivatives is then applied. This rule states that for two functions $u(x)$ and $v(x)$:

   $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{v(x)\frac{d}{dx}u(x)-u(x)\frac{d}{dx}v(x)}{[v(x){]}^2}$

   In this case, $u(x)=\sin (x)$ and $v(x)=\cos (x)$. Applying the quotient rule gives:

   $\frac{d}{dx}\tan (x)=\frac{\cos (x)\cos (x)-\sin (x)(-\sin (x))}{{\cos }^2(x)}$

4. The numerator is simplified using the standard trigonometric identity ${\cos }^2(x)+{\sin }^2(x)=1$:

   $\frac{d}{dx}\tan (x)=\frac{{\cos }^2(x)+{\sin }^2(x)}{{\cos }^2(x)}$

5. Using the identity from step 4, the numerator simplifies to 1:

   $\frac{d}{dx}\tan (x)=\frac{1}{{\cos }^2(x)}$

6. The proof is valid only when $\cos (x)\ne 0$, as division by zero is undefined.

7. Finally, the result follows from the fact that $\sec (x)=\frac{1}{\cos (x)}$ (the secant is the reciprocal of the cosine).

Therefore, the derivative of $\tan (x)$ is ${\sec }^2(x)$.

Proof #2

Explanation

1. The proof begins with the definition of the derivative of a real function at a point. In this case, it’s the derivative of tangent with respect to $x$, which is the limit as $h$ approaches $0$ of $\frac{\tan (x+h)-\tan x}{h}$.

2. The next step uses the trigonometric identity for the tangent of a sum: $\tan (A+B)=\frac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}$. Here, $A$ is $x$ and $B$ is $h$. Applying this identity to $\tan (x+h)$, we get: $\frac{\tan (x)+\tan (h)}{1-\tan (x)\tan (h)}$.

3. The numerator is then expanded by adding and subtracting $\tan (x)$: $\frac{\tan (x)+\tan (h)-\tan (x)+{\tan }^2(x)\tan (h)}{1-\tan (x)\tan (h)}$.

4. The numerator is factored and the denominator is multiplied by $h$: $\frac{\tan (h)+{\tan }^2(x)\tan (h)}{h(1-\tan (x)\tan (h))}$.

5. The product rule for limits is applied, splitting the limit into the product of two limits: ${lim}_{h\to 0}\frac{1+{\tan }^2(x)}{1-\tan (x)\tan (h)}·{lim}_{h\to 0}\frac{\tan (h)}{h}$.

6. The second limit is a standard limit: ${lim}_{h\to 0}\frac{\tan (h)}{h}=1$. In the first limit, $\tan (0)=0$, so as $h\to 0$, $\tan (h)\to 0$. Thus, the first limit evaluates to $\frac{1+{\tan }^2(x)}{1-\tan (x)\tan 0}=1+{\tan }^2(x)$.

7. The result is simplified using the trigonometric identity $1+{\tan }^2(x)={\sec }^2(x)$.

8. Finally, the result is expressed in terms of cosine using the identity $\sec (x)=\frac{1}{\cos (x)}$, provided $\cos (x)\ne 0$.

Therefore, the derivative of $\tan (x)$ with respect to $x$ is $\boxed{{\sec }^2(x)}$ or $\frac{1}{{\cos }^2(x)}$, provided $\cos (x)\ne 0$.