Q11 Ex 3.7 Thomas Calculus Twelfth Edition | Answer Key - Edubirdie

University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2024
Views:

26

Pages:

2
Author:

Shahbaz A.

Q11 Ex 3.7 THOMAS CALCULUS Twelfth Edition shahbaz ahmed July 2024 Use implicit differentiation to find dy dx xm y n = (x + y)m+n Q11 Solution xm y n = (x + y)m+n ln(xm y n ) = ln(x + y)m+n ln xm + ln y n = (m + n) ln(x + y) m ln x + n ln y = (m + n) ln(x + y) d(m ln x+n ln y) dx d(m ln x) dx + = d((m+n) ln(x+y)) dx d(n ln y) dx = (m + n) d(ln(x+y)) dx x) d(ln y) 1 d(x+y) m d(ln dx + n dx = (m + n) x+y dx m x y) dy 1 dx + n d(ln dy dx = (m + n) x+y [ dx + m x + n dy y dx 1 = (m + n) x+y [1 + dy dx ] 1 dy dx ] m x + n dy y dx = m+n x+y [1 m x + n dy y dx = m+n x+y n dy y dx − [ ny − m+n dy x+y dx m+n dy x+y ] dx = = + + dy dx ] m+n dy x+y dx m+n x+y m+n x+y − − m x m x dy ] dx = [ nx+ny−(m+n)y y(x+y) (m+n)x−m(x+y) x(x+y) dy [ nx+ny−my−ny ] dx = y(x+y) (mx+nx−mx−my) x(x+y) dy [ nx−my y(x+y) ] dx = (nx−my) x(x+y) Since (nx − my) ̸= 0 and similarly (x + y) ̸= 0 dy [ y1 ] dx = dy dx = 1 x y x 2

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