y = [x + (x + sin²x)³]^4 I ended up getting 4[x + x(x+sin²x)³]³ * 3x(sin²x)² * 2cosx Am I right???! If not, can someone seriously walk me through this? This was a really messy problem for me.
Last edited by AntiSandman; Oct 12th 2008 at 08:27 PM.
Follow Math Help Forum on Facebook and Google+
Originally Posted by AntiSandman y = [x + x(sin²x)³]^4 I ended up getting 4[x + x(x+sin²x)³]³ * 3x(sin²x)² * 2cosx Am I right???! If not, can someone seriously walk me through this? This was a really messy problem for me. nope, not right. how about i give you a few hints and you try again and see where we get to find the derivative of you need to use the product rule when you are differentiating the part, you need to use the chain rule. it would be easier to write as by the way
y = [x + (x + sin²x)³]^4 I made a mistake, sorry, that's the real problem. Am I still right? (editted the above post too)
Originally Posted by AntiSandman y = [x + (x + sin²x)³]^4 I made a mistake, sorry, that's the real problem. Am I still right? (editted the above post too) nope, still note right. you need the chain rule to differentiate . plus, what about the derivative of the lone x?
Is this it? 4[x + x(x+sin²x)³]³ * [1 + 3(x + sin²x)² * (1 + 2cosxsinx)]
Originally Posted by AntiSandman Is this it? 4[x + x(x+sin²x)³]³ * [1 + 3(x + sin²x)² * (1 + 2cosxsinx)] yup, that's it you can write 2cos(x)sin(x) as sin(2x) though. maybe simplify even more if you are so inclined (i wouldn't be ) wait, you have an extra x. the one in red shouldn't be there
Originally Posted by Jhevon yup, that's it you can write 2cos(x)sin(x) as sin(2x) though. maybe simplify even more if you are so inclined (i wouldn't be ) wait, you have an extra x. the one in red shouldn't be there Thanks so much, I seriously appreciate you helping me out so late at night.
View Tag Cloud