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Calculus help?
Find derivative of cosh(ln(9x^3)).
pls explain thanks!!
2 Answers
- ted sLv 73 years ago
comment : cosh w = [ e^w + e^(-w) ] / 2.....thus dcosh ( ln(9x³) / dx = ( 27 / 2 ) x² - 1 / ( 6 x^4)
- ?Lv 53 years ago
OK, here you go sweetheart :)
Just use the chain rule and you should be fine!
Let y = cosh(u), with u = ln(9x³)
dy/du = sinh(u) and du/dx = 3/x
Chain Rule :
dy/dx = (dy/du) × (du/dx)
dy/dx = sinh(u) × (3/x)
dy/dx = (sinh(ln(9x³))) × (3/x)
Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
P.S. (Don't forget to vote me best answer as being the first to correctly answer your question!)
