The **derivative of tan(x)** is sec²x. Let’s prove this!

**?Prior knowledge**: You need to know these formulas (*Trigonometric Identity of tan(x), Derivative of sin(x), and Derivative of cos(x)*) before learning the derivative of tan(x):

[latexpage]

\[ tanx=\frac{sinx}{cosx} \]

\[ \frac{d}{dx}sinx=cosx \]

\[ \frac{d}{dx}cosx=-sinx \]

We will use the **quotient rule** in order to calculate (tanx)’. So, it is useful to learn what the quotient rule is. Because we know that:

\[ tan(x)=\frac{sinx}{cosx}\]

**?Keep in mind: ** If two functions u = f(x) and v = g(x) are given and these functions are differentiable, we can say that:

\[ (\frac{u}{v})’ = \frac{u’.v – v’.u}{v^2}\]

The derivative of a quotient is the denominator times the **derivative of the numerator** **minus** the numerator **times the derivative of the denominator**,** all divided by **the **square** of the denominator.

We can continue with making some calculations in order to find (tanx)’

\[ (tanx)’=(\frac{sinx}{cosx})’\]

\[ (tanx)’=\frac{cosx.cosx – sinx.(-sinx)}{cos^2x}\]

Lets simplify the numerator of this function:

\[ (tanx)’=\frac{cos^2x + sin^2x}{cos^2x}\]

Here is a basic trigonometry equation hidden! **cos ^{2}x + sin^{2}x = 1**. So,

\[ (tanx)’=\frac{1}{cos^2x} = sec^2x \]

- Finally, we proved the
**derivative of tanx**is sec²x. ?

Let’s continue with an example. Below, there is an equation but it does not contain tan. We will try to take the derivative of this expression. We will use both the basic rules of the derivative and the trigonometric identities we mentioned above.

\[ f(x)=\frac{sinx + sin3x + sin5x}{cosx + cos3x + cos5x} \Rightarrow f'(x)=?\]

First of all, f(x) equals ** tan3x** because of a trigonometric identity which is the sum of the product formula. The derivative of tan3x is asked in the question. Use the chain rule, that’s all! ✅

The answer is** (tan3x)’** = **3. sec ^{2}3x**