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Differentiate the function with respect to x. โˆš(๐‘ฅโˆ’1)โข(๐‘ฅโˆ’2)/(๐‘ฅโˆ’3)โข(๐‘ฅโˆ’4)โข(๐‘ฅโˆ’5) - Mathematics

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Question

Differentiate the function with respect to x.

`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`

Sum
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Solution

Let, y = `sqrt(((x - 1)(x - 2))/((x - 3)(x - 4)(x - 5)))`  ...(1)

or, y = `[((x - 1)(x - 2))/((x - 3)(x - 4)(x - 5))]^(1/2)`

Taking logarithm of both sides,

log y = `1/2 ((x - 1)(x - 2))/((x - 3)(x - 4)(x - 5))`   ...[โˆต log mn = n log m]

Or log y = `1/2 log (x - 1) (x - 2) - 1/2 log (x - 3) (x - 4) (x - 5)    ...[โˆต log m/n = log m - log n]`

= `1/2 [log (x- 1) + log (x - 2)] - 1/2 [log (x - 3) + log (x - 4) + log (x - 5)]`   ...[โˆต log m . n = log m + log n]

Representing both sides by x,

`1/y dy/dx = 1/2 [d/dx log (x - 1) + d/dx log (x - 2)] - 1/2 [d/dx log (x - 3) + d/dx log (x - 4) + d/dx log (x - 5)]`

= `1/2 y [1/(x - 1) d/dx (x - 1) + 1/(x - 2) d/dx (x - 2)] - 1/2 y [1/(x - 3) d/dx (x - 3) + 1/(x - 4) d/dx (x - 4) + 1/(x -  5) d/dx (x - 5)]`

= `1/2 y [1/(x - 1) + 1/(x - 2)] - 1/2 y [1/(x - 3) + 1/(x - 4) + 1/(x - 5)]`

= `1/2 y [1/(x - 1) + 1/(x - 2) - 1/(x - 3) - 1/(x - 4) - 1/(x - 5)]`

Putting the value of y in equation (1),

`dy/dx =  1/2 sqrt(((x - 1)(x - 2))/((x - 3)(x - 4)(x - 5))) [1/(x - 1) + 1/(x - 2) - 1/(x - 3) - 1/(x - 4) - 1/(x - 5)]`

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Chapter 5: Continuity and Differentiability - Exercise 5.5 [Page 178]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.5 | Q 2 | Page 178

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