Question

cos(ax+b) differentiate the function by first principle of differentiate

Answer 1

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Answer 2

Complete Question

Differentiate the function cos(ax + b) by first principle of differentiation

Answer

The derivative of cos(ax + b) is -asin(ax + b)

Given

cos(ax + b)

To Find

First-order derivative by the first principle differentiating method

Solution

The formula for Differentiating any function f(x) with respect to x using the first principle method is:-

$f'(x)= \lim_{h\to 0}\frac{ f(x+h)- f(x)}{h}$                                              [1]

where f'(x) is the first-order derivative of f(x)

Here,

f(x) =                                                                     [2]

f(x+h) = cos[a(x+h) + b]

= cos[ax + ah + b]                                                                  [3]

From equations [1], [2], and [3] we get

$f'(x)= \lim_{h\to 0} \frac{cos[ax + ah + b]- cos(ax + b)}{h}$

$= \lim_{h\to 0} \frac{cos[ax +b+ ah ]- cos(ax + b)}{h}$                                               [4]

cos(C + D) = cosCcosD - sinCsinD

Using this in equation [4]by considering ax + b as one term and ah as another  gives us

$= \lim_{h\to 0} \frac{cos(ax +b)cos(ah) -sin(ax+b)sin(ah) - cos(ax + b)}{h}$

We will now separate the sin term out.

lim(c + d) = lim(c) + lim(d)

$= \lim_{h\to 0} \frac{cos(ax +b)cos(ah) - cos(ax + b)}{h} -\lim_{h\to 0}\frac {sin(ax+b)sin(ah)}{h}$

taking cos(ax+b) common and sin(ax + b) out of the limit we get

$= \lim_{h\to 0} \frac{cos(ax +b)[cos(ah) - 1]}{h} -sin(ax+b)\lim_{h\to 0}\frac {sin(ah)}{h}$

Now we will multiply a with the numerator and denominator of the second limit and use the formula

1 - cosy = 2sin²(y/2) for the first term

$=- \lim_{h\to 0} \frac{cos(ax +b)2sin^2(ah/2)}{h} -sin(ax+b)\lim_{h\to 0}\frac {asin(ah)}{ah}$

Manipulating the first term we get

$=- cos(ax +b)\lim_{h\to 0} \frac{asin^2(ah/2)}{ah/2} -asin(ax+b)\lim_{h\to 0}\frac {sin(ah)}{ah}$

Now $\lim_{h \to 0} sinh/h = 1$

using this we get

$=- cos(ax +b)\lim_{h\to 0} asin(ah/2) -asin(ax+b)$

putting the value of h gives us

$=0 -asin(ax+b)$

= -asin(ax + b)

Hence the derivative of cos(ax + b) is -asin(ax + b)

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