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Find the derivative of the inverse of the following functions, and also find their value at the points indicated against them. y = sin(x – 2) + x2 - Mathematics and Statistics

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प्रश्न

Find the derivative of the inverse of the following functions, and also find their value at the points indicated against them. y = sin(x – 2) + x2 

बेरीज

उत्तर

y = sin(x – 2) + x2 
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[sin(x - 2) + x^2]`

= `"d"/"dx"[sin(x - 2)] + "d"/"dx"(x^2)`

= `cos(x - 2)."d"/"dx"(x - 2) + 2x`
= cos(x – 2).(1 – 0) + 2x
= cos(x – 2) + 2x
The derivative of inverse function of y = f(x) is given by
`"dx"/"dy" = (1)/(("dy"/"dx")`

= `(1)/(cos(x - 2) + 2x)`

At x = `2,"dx"/"dy"`

= `((1)/([cos(x - 2) + 2x]))_("at"  x = 2)`

= `(1)/(cos0 + 2(2)`

= `(1)/(1 + 4)`

= `(1)/(5)`.

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Derivatives of Inverse Functions
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Differentiation - Exercise 1.2 [पृष्ठ २९]

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