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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

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

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

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