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Find the derivative of the following w. r. t. x. at the point indicated against them by using method of first principle: cos x at x = 5π4 - Mathematics and Statistics

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प्रश्न

Find the derivative of the following w. r. t. x. at the point indicated against them by using method of first principle:

cos x at x = `(5pi)/4`

बेरीज
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उत्तर

Let f(x) = cos x

∴ `"f"((5pi)/4) = cos ((5pi)/4)`

`"f"((5pi)/4 + "h") = cos ((5pi)/4 + "h")`

∴ `"f"((5pi)/4 + "h") - "f"((5pi)/4)`

= `cos((5pi)/4 + "h") - cos ((5pi)/4)`

= `-2sin [((5pi)/4 + "h" + (5pi)/4)/2] sin [((5pi)/4 + "h" - (5pi)/4)/2]`

= `-2 sin [((5pi)/2 + "h")/2] sin ("h"/2)`

By definition,

`"f'"((5pi)/4) =  lim_("h" -> 0) ("f"((5pi)/4 + "h") - "f"((5pi)/4))/"h"`

= `lim_("h" -> 0) (-2sin [((5pi)/2 + "h")/2] sin("h"/2))/"h"`

= `lim_("h" -> 0) [-2 sin [((5pi)/2 + "h")/2] (sin("h"/2))/(("h"/2))] xx 1/2`

= `-[lim_("h" -> 0) sin [((5pi)/2 + "h")/2] ] xx [lim_("h" -> 0) (sin("h"/2))/("h"/2)]`

= `-sin [((5pi)/2 + 0)/2] xx 1  ...[because "h" -> 0, "h"/2 -> 0  "and" lim_(theta -> 0) sintheta/theta = 1]`

= `- sin ((5pi)/4)`

= `- sin (pi + pi/4)`

= `-(- sin  pi/4)`

= `1/sqrt(2)`

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Definition of Derivative and Differentiability
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differentiation - Exercise 9.1 [पृष्ठ १८७]

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