Notes On Derivatives of Trigonometric Functions - CBSE Class 11 Maths
Derivatives of trigonometric functions are unlike the derivatives of algebraic functions. Derivatives of some trigonometric functions using the first principle method: sin x, cos x, tan x Derivative of Sin x f(x) = sin x f '(x) = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{f(x+h)-f(x)}}{\text{h}}$   f '(x) =            = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{2 cos(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$   Since, sin A -sin B = 2 cos(A+B / 2) sin(A-B / 2)          = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{2 cos(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$   As h → 0, we have h/2 → 0           = $\underset{h\to \text{0}}{\text{lim}}\text{cos(}\frac{\text{2x+h}}{\text{2}}\text{)}\text{.}\underset{\mathrm{h/2}\to \text{0}}{\text{lim}}\frac{\text{sin(}\frac{\text{h}}{\text{2}}\text{)}}{\frac{\text{h}}{\text{2}}}$$\text{}$                  = cos x × 1 fI(x) = cos x Therefore, the derivative of sin x is cos x. Derivative of Cos x f(x) = cos x            = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{- 2 sin(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$ Since cos A - cos B = -2 sin(A+B /2)sin(A-B /2) as h → 0, we have h/2 → 0 = $\underset{h\to \text{0}}{\text{lim}}\text{- sin(}\frac{\text{2x+h}}{\text{2}}\text{)}\text{.}\underset{\mathrm{h/2}\to \text{0}}{\text{lim}}\frac{\text{sin(}\frac{\text{h}}{\text{2}}\text{)}}{\frac{\text{h}}{\text{2}}}$$\text{}$ = (- sin x) × 1    (∴ $\underset{x\to \text{0}}{\text{lim}}\frac{\text{sin x}}{\text{x}}$ = 1) fI(x) = -sin x Therefore, the derivative of cos x is - sin x. Derivative of Tan x f(x) = tan x f '(x) = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{tan(x+h)- tanx}}{\text{h}}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h)}}{\text{cos(x+h)}}\text{-}\frac{\text{sin x}}{\text{cos x}}\text{]}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h).cos x - sin x. cos(x+h)}}{\text{cos(x+h)}}\text{]}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h-x)}}{\text{cos(x+h).cos x}}\text{]}$    [Since sin(A - B) = sin A cos B - cos A sin B]          = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(h)}}{\text{cos(x+h).cos x}}\text{]}$   =   = 1 × [1/(cos x cos x)] = 1 × [1/(cos2 x)] = sec2 x The derivative of tan x is secant2 x.

#### Summary

Derivatives of trigonometric functions are unlike the derivatives of algebraic functions. Derivatives of some trigonometric functions using the first principle method: sin x, cos x, tan x Derivative of Sin x f(x) = sin x f '(x) = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{f(x+h)-f(x)}}{\text{h}}$   f '(x) =            = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{2 cos(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$   Since, sin A -sin B = 2 cos(A+B / 2) sin(A-B / 2)          = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{2 cos(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$   As h → 0, we have h/2 → 0           = $\underset{h\to \text{0}}{\text{lim}}\text{cos(}\frac{\text{2x+h}}{\text{2}}\text{)}\text{.}\underset{\mathrm{h/2}\to \text{0}}{\text{lim}}\frac{\text{sin(}\frac{\text{h}}{\text{2}}\text{)}}{\frac{\text{h}}{\text{2}}}$$\text{}$                  = cos x × 1 fI(x) = cos x Therefore, the derivative of sin x is cos x. Derivative of Cos x f(x) = cos x            = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{- 2 sin(}\frac{\text{2x+h}}{\text{2}}\text{)sin(}\frac{\text{h}}{\text{2}}\text{)}}{\text{h}}$ Since cos A - cos B = -2 sin(A+B /2)sin(A-B /2) as h → 0, we have h/2 → 0 = $\underset{h\to \text{0}}{\text{lim}}\text{- sin(}\frac{\text{2x+h}}{\text{2}}\text{)}\text{.}\underset{\mathrm{h/2}\to \text{0}}{\text{lim}}\frac{\text{sin(}\frac{\text{h}}{\text{2}}\text{)}}{\frac{\text{h}}{\text{2}}}$$\text{}$ = (- sin x) × 1    (∴ $\underset{x\to \text{0}}{\text{lim}}\frac{\text{sin x}}{\text{x}}$ = 1) fI(x) = -sin x Therefore, the derivative of cos x is - sin x. Derivative of Tan x f(x) = tan x f '(x) = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{tan(x+h)- tanx}}{\text{h}}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h)}}{\text{cos(x+h)}}\text{-}\frac{\text{sin x}}{\text{cos x}}\text{]}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h).cos x - sin x. cos(x+h)}}{\text{cos(x+h)}}\text{]}$ = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(x+h-x)}}{\text{cos(x+h).cos x}}\text{]}$    [Since sin(A - B) = sin A cos B - cos A sin B]          = $\underset{h\to \text{0}}{\text{lim}}\frac{\text{1}}{\text{h}}$$\text{[}\frac{\text{sin(h)}}{\text{cos(x+h).cos x}}\text{]}$   =   = 1 × [1/(cos x cos x)] = 1 × [1/(cos2 x)] = sec2 x The derivative of tan x is secant2 x.

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