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Tamil Nadu Board of Secondary EducationHSC Science Class 11

Differentiate the following:y = (2x – 5)4 (8x2 – 5)–3 - Mathematics

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Question

Differentiate the following:
y = (2x – 5)4 (8x2 – 5)–3 

Sum
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Solution

y = (2x – 5)4 (8x2 – 5)–3 

`("d"y)/("d"x) = (2x - 5)^4 xx "d"/("d"x) (8x^2 - 5)^(-3) + (8x^2 - 5)^(-3) xx "d"/("d"x) (2x - 5)^4`  .......(By product rule)

`("d"y)/("d"x) = (2x - 5)^4 xx - 3(8x^2 - 5)^(- 3 - 1) (8 xx 2x - 0) + (8x^2 - 5)^(-3) xx 4(2x - 5)^(4 - 1) (2 xx 1 - 0)`

`("d"y)/("d"x) = - 3(2x - 5)^4 (8x^2 - 5)^(-4) xx 16x + (8x^2 - 5)^(-3) xx 4(2x - 5)^3 (2 xx 1)`

`("d"y)/("d"x) = - 48x (2x - 5)^4 (8x^2 - 5)^(- 4) + 8(8x^2 - 5)^(- 3) (2x - 5)^3`

= `-48x (2x - 5)^4 xx 1/(8x^2 - 5)^4 + (8(2x - 5)^3)/(8x^2 - 5)^3`

= `(8(2x - 5)^3)/(8x^2 - 5)^3 [1 - (6x(2x - 5))/(8x^2 - 5)]`

= `(8(2x - 5)^3)/(8x^2 - 5)^3 xx [(8x^2 - 5 - 12x^2 + 30x)/(8x^2 - 5)]`

= `(8(2x - 5)^3)/(8x^2 - 5)^3 xx [- 4x^2 + 30x - 5]`

`("d"y)/("d"x) = (8(2x - 5)^3)/(8x^2 - 5)^4 xx (30x - 4x^2 - 5)`

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Chapter 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.3 [Page 163]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.3 | Q 13 | Page 163

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