what is the derivative of cosx^2
D Daniel_Feldman Full Member Joined Sep 30, 2005 Messages 252 Jan 21, 2006 #2 jeflow said: what is the derivative of cosx^2 Click to expand... If you want help, you should be clear as to what the problem is. Is this (cosx)^2 or cos(x^2)> For the first one f(x)=(cosx)^2 Using power and chain rules: f'(x)=2(cosx)(-sinx)=-2sinxcosx=-sin(2x) If f(x)=cos(x^2) We use chain and power rules: f'(x)=-sin(x^2)(2x)=-2xsin(x^2)
jeflow said: what is the derivative of cosx^2 Click to expand... If you want help, you should be clear as to what the problem is. Is this (cosx)^2 or cos(x^2)> For the first one f(x)=(cosx)^2 Using power and chain rules: f'(x)=2(cosx)(-sinx)=-2sinxcosx=-sin(2x) If f(x)=cos(x^2) We use chain and power rules: f'(x)=-sin(x^2)(2x)=-2xsin(x^2)
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jan 21, 2006 #3 It's not that bad to do on your own. Assuming you mean cos2(x)\displaystyle cos^{2}(x)cos2(x): Let u=cos(x)\displaystyle u=cos(x)u=cos(x), then du=−sin(x)dx\displaystyle du=-sin(x)dxdu=−sin(x)dx ddu[u2]=2ududx\displaystyle \frac{d}{du}[u^{2}]=2u\frac{du}{dx}dud[u2]=2udxdu =2cos(x)(−sin(x))=−2cos(x)sin(x)\displaystyle =2cos(x)(-sin(x))=-2cos(x)sin(x)=2cos(x)(−sin(x))=−2cos(x)sin(x)
It's not that bad to do on your own. Assuming you mean cos2(x)\displaystyle cos^{2}(x)cos2(x): Let u=cos(x)\displaystyle u=cos(x)u=cos(x), then du=−sin(x)dx\displaystyle du=-sin(x)dxdu=−sin(x)dx ddu[u2]=2ududx\displaystyle \frac{d}{du}[u^{2}]=2u\frac{du}{dx}dud[u2]=2udxdu =2cos(x)(−sin(x))=−2cos(x)sin(x)\displaystyle =2cos(x)(-sin(x))=-2cos(x)sin(x)=2cos(x)(−sin(x))=−2cos(x)sin(x)
