Asked by Kelly
I've done everything I can think of to solve this, but I have no idea. Find the derivative: y = sq rt{11x + sqrt[11x + sqrt(11x)]}
Answers
Answered by
Steve
I think you mean
y = √(11x + √(11x+√(11x)))
y^2 = 11x + √(11x+√(11x))
If u^2 = 11x,
u' = 11/(2√(11x))
y^2 = u^2 + √(u^2+u)
2y y' = 2u u' + (2u+1)u'/(2√(u^2+u))
2y y' = u' (4u+1)/(2√(u^2+u))
Now substitute back in for y, u, u' and you will get the mess shown here:
http://www.wolframalpha.com/input/?i=derivative+%E2%88%9A%2811x+%2B+%E2%88%9A%2811x%2B%E2%88%9A%2811x%29%29%29
Problems like this are kinda fun, but they really don't help you learn any calculus -- just keeping track of algebraic details.
y = √(11x + √(11x+√(11x)))
y^2 = 11x + √(11x+√(11x))
If u^2 = 11x,
u' = 11/(2√(11x))
y^2 = u^2 + √(u^2+u)
2y y' = 2u u' + (2u+1)u'/(2√(u^2+u))
2y y' = u' (4u+1)/(2√(u^2+u))
Now substitute back in for y, u, u' and you will get the mess shown here:
http://www.wolframalpha.com/input/?i=derivative+%E2%88%9A%2811x+%2B+%E2%88%9A%2811x%2B%E2%88%9A%2811x%29%29%29
Problems like this are kinda fun, but they really don't help you learn any calculus -- just keeping track of algebraic details.
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