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**Derivatives of Trigonometric Functions:**

Trigonometric and inverse trigonometric functions are differentiable at each point of its domain.

- \(\frac{d}{dx} \sin x = \cos x \)
- \(\frac{d}{dx} \cos x = – \sin x \)
- \(\frac{d}{dx} \tan x = \sec^2 x \)
- \(\frac{d}{dx} \cot x = – cosec^2 x \)
- \(\frac{d}{dx} \sec x = \sec x . \tan x\)
- \(\frac{d}{dx} cosec x = – cosec x. \cot x\)

**Note:** remember that the trigonometric functions which are starts with **c** then their derivate should come with the minus(-) sign preceding the derivative of the functions and also applicable in case of **the inverse trigonometric functions** Ex: \(\frac{d}{dx} \cos x = – \sin x\)

**Derivatives of Inverse Trigonometric Functions:**

- \(\frac{d}{dx} \sin^{-1}x = \frac{1}{\sqrt{1- x^2}}, (|x|<1)\)
- \(\frac{d}{dx} \cos^{-1}x = – \frac{1}{\sqrt{1- x^2}}, (|x|<1)\)
- \(\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}\), \(x \in R \)
- \(\frac{d}{dx} \cot^{-1}x = -\frac{1}{1+x^2}\), \(x \in R \)
- \(\frac{d}{dx} \sec^{-1}x = \frac{1}{x \sqrt{x^2-1}}, (|x|>1)\)
- \(\frac{d}{dx} cosec^{-1}x = -\frac{1}{x \sqrt{x^2- 1}}, (|x|>1)\)

**Derivatives of Exponential and Logarithmic Functions:**

- \(\frac{d}{dx}e^x = e^x\)
- \(\frac{d}{dx}a^x = a^x \log_ea, a>0, a \neq 1\)
- \(\frac{d}{dx}\log_ex = \frac{1}{x}, x>0\)
- \(\frac{d}{dx} \log_ax = \frac{1}{xlog_ea} = \frac{log_ae}{x}\)

The logarithmic function is differentiable at each point of its domain. \(a^x\) is differentiable at each \(x \in , (a >0, a\neq 1)\)

**Derivatives of Hyperbolic Functions:**

- \(\frac{d}{dx} \sin h x = \cos h x \)
- \(\frac{d}{dx} \cos h x = – \sin h x \)
- \(\frac{d}{dx} \tan h x = \sec h^2 x \)
- \(\frac{d}{dx} \cot h x = – cosec h^2 x \)
- \(\frac{d}{dx} \sec h x = \sec h x . \tan h x\)
- \(\frac{d}{dx} cosec h x = – cosec h x. \cot h x\)

**Parametric Differentiation**

If x = f(t), y =g(t) are both the derivable functions of ‘t’ then

\(\frac{d}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}, (f'(t) \neq 0)\)

**Derivative of Infinite series:**

- If \(y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x)+ \cdots + \infty}}}\) then \(y = \sqrt{f(x)+y}\)

Squaring:

\(y^2 = f(x) +y \\y^2-y = f(x)\)

Differentiating both sides w.r.t. x, we get

\((2y-1) \frac{dy}{dx} = f'(x)\)

For example let take one trigonometric function:

\(y = \sqrt{\sin x+ \sqrt{ \sin x + \sqrt{\sin x+ \cdots + \infty}}}\) then \((2y -1) \frac{dy}{dx} = \cos x\)

**Note: **do like this for each and every functions to get derivative of the functions.