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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
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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