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Chapter 6: Problem 4

Find the derivative of \(y=f(x)=\sin x\), using first principles.

Short Answer

Expert verified
The derivative of \(y = \sin x\) is \(f'(x) = \cos x\).

Step by step solution

01

Definition of Derivative using First Principles

According to the definition of derivative using first principles, the derivative of a function \(f(x)\) is given by: \[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\] Substitute \(f(x) = \sin x\) into the above definition.
02

Substituting sine function into definition

Substituting \(f(x) = \sin x\) into the definition we get: \[f'(x) = \lim_{h\to 0} \frac{\sin(x+h)-\sin x}{h}\]
03

Applying the trigonometric identity

Use the sine addition formula \(\sin(a+b) = \sin a \cos b + \cos a \sin b\), where a = x and b = h in \(\sin(x+h)\): \[f'(x) = \lim_{h\to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\]
04

Rearranging the expression

Rearrange expression to separate terms that contain \(h\): \[f'(x) = \lim_{h\to 0} (\sin x (\frac{\cos h - 1}{h}) + \cos x \frac{\sin h}{h}) \]
05

Applying Limit Properties

Apply limit properties and remember that \(\lim_{h\to 0} \frac{\sin h}{h} = 1\) and \(\lim_{h\to 0} \frac{\cos h - 1}{h} = 0\) : \[f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x\]

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Most popular questions from this chapter

If \(\sqrt{x^{2}+y^{2}}=a e^{\tan ^{-1}} x\), where \(a>0, y(0) \neq 0\) then find the value of \(y^{\prime \prime}(0)\).

For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)

If \(y=\sin ^{-1} x\), prove that \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\)

Find \(\frac{d y}{d x}\), if \(y=(\tan x)^{\cot x}+(\cot x)^{\tan x}\)

If \(y=x+\tan x\), prove that \(\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0\)
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