K KK88 May 2010 6 1 May 10, 2010 #1 Hi I have a few calculus questions, could you please explain how to get the final answer too, thanks. Derivative of : dy/dx here dy/dx here dy/dt here Last edited: May 10, 2010

Hi I have a few calculus questions, could you please explain how to get the final answer too, thanks. Derivative of : dy/dx here dy/dx here dy/dt here

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #2 KK88 said: Hi I have a few calculus questions, could you please explain how to get the final answer too, thanks. Derivative of : Chain rule of the chain rule with a product rule chain rule dy/dx here chain rule dy/dx here product rule with chain dy/dt here chain Click to expand... You haven't attempted much so start by trying that.

KK88 said: Hi I have a few calculus questions, could you please explain how to get the final answer too, thanks. Derivative of : Chain rule of the chain rule with a product rule chain rule dy/dx here chain rule dy/dx here product rule with chain dy/dt here chain Click to expand... You haven't attempted much so start by trying that.

K KK88 May 2010 6 1 May 10, 2010 #3 Ok I figured out the middle three but I cannot quite get the first one and last one

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #4 \(\displaystyle (cos\big(e^{x^2cos(x)}\big))^{\frac{1}{2}}\) \(\displaystyle \frac{1}{2}(cos\big(e^{x^2cos(x)}\big))^{\frac{-1}{2}}(-sin\big(e^{x^2cos(x)}\big))(e^{x^2cos(x)}(2xcos(x)-x^2sin(x)))\) I am not going to simplify that. Reactions: KK88

\(\displaystyle (cos\big(e^{x^2cos(x)}\big))^{\frac{1}{2}}\) \(\displaystyle \frac{1}{2}(cos\big(e^{x^2cos(x)}\big))^{\frac{-1}{2}}(-sin\big(e^{x^2cos(x)}\big))(e^{x^2cos(x)}(2xcos(x)-x^2sin(x)))\) I am not going to simplify that.

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #5 \(\displaystyle e^{2sin(8t)}\) \(\displaystyle 8e^{2sin(8t)}2cos(8t)=16e^{2sin(8t)}cos(8t)\) Reactions: KK88

K KK88 May 2010 6 1 May 10, 2010 #6 Thanks for the responses. The second one is correct however I still cant get the first one right, I input this: Reactions: Rapha

Thanks for the responses. The second one is correct however I still cant get the first one right, I input this:

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #7 The first one is correct too. Here is the maple output. Reactions: KK88

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #8 \(\displaystyle \frac{1}{2}(cos\big(e^{x^2cos(x)}\big))^{\frac{-1}{2}}(-sin\big(e^{x^2cos(x)}\big))(e^{x^2cos(x)}(2xcos(x)-x^2sin(x)))\) You need to have this quantity in parenthesis. (2xcos(x)-x^2sin(x)) Reactions: KK88

\(\displaystyle \frac{1}{2}(cos\big(e^{x^2cos(x)}\big))^{\frac{-1}{2}}(-sin\big(e^{x^2cos(x)}\big))(e^{x^2cos(x)}(2xcos(x)-x^2sin(x)))\) You need to have this quantity in parenthesis. (2xcos(x)-x^2sin(x))