Answer to Question #92552 in Calculus for Francis

Question

1. y= cosh( sin 2/ x squared)
2. [ coth raised to -1 ( sinh(2 raised to 3x))] raised to cube
3. y= x raised to ln (x+2)
4. y= log with base of 4 ( x squared + 3 ) squared square root of 2x -4 / ( x-4 ) raise to 4

Answer 1

$1) (cosh(sin^2(\cfrac{2}{x})))' = sinh(sin^2(2/x))2sin(\cfrac{2}{x})cos(\cfrac{2}{x})(\cfrac{-2}{x^2})$

$2)(\cfrac{1}{coth^3(sinh(2^{3x}))})'=\cfrac{-3}{coth^4(sinh(2^{3x}))}\cfrac{-3ln(2)2^{3x}cosh(2^{3x})}{sinh^2(sinh(2^{3x}))}=\cfrac{9ln(2)2^{3x}cosh(2^{3x})}{coth^4(sinh(2^{3x}))sinh^2(sinh(2^{3x}))}$$3)(x^{ln(x+2)})'=x^{ln(x+2)}(ln(x^{ln(x+2)})'=x^{ln(x+2)}(ln(x+2)ln(x))'=x^{ln(x+2)}(\cfrac{ln(x+2)}{x}+\cfrac{lnx}{x+2})$

$4)(\log_4(x^2+3)\cfrac{\sqrt{2x-4}}{(x-4)^4})'=(\log_4(x^2+3)\sqrt{2x-4})'\cfrac{1}{(x-4)^4}+ (\log_4(x^2+3)\sqrt{2x-4})(\cfrac{1}{(x-4)^4})'=-4\log_4(x^2+3)\cfrac{\sqrt{2x-4}}{(x-4)^5}+(\log_4(x^2+3)\cfrac{1}{(x-4)^4\sqrt{2x-4}})+\cfrac{\sqrt{2x-4}}{(x-4)^4}\cfrac{2x}{(x^2+3)ln4}$