2017年9月15日 · We have: y = xsinx Which is the product of two functions, and so we apply the Product Rule for Differentiation: d dx (uv) = u dv dx + du dx v, or, (uv)' = (du)v +u(dv) I was taught to remember the rule in words; " The first times the derivative of the second plus the derivative of the first times the second ". So with y = xsinx; {Let u = …

2018年4月4日 · General solution for your differential equation is: y=(x-1)cosx-sinx+C . y'=dy/dx=sinx-xsinx To find y, we have to take the integral of y': y=int(sinx-xsinx)dx y=intsinxdx-intxsinxdx=-cosx-I I=intxsinxdx The argument of the integral is product of two functions. As such, we will use integration by parts: u=x, and dv=sinxdx du=dx, and v=-cosx intudv=uv-intvdu intxsinxdx=-xcosx-int-cosxdx= -xcosx ...

2017年3月9日 · Reminder ∙ d dx (sinx) = cosx and d dx (cosx) = − sinx to differentiate xsinx use the product rule Given f (x) = g(x)h(x) then

2016年12月11日 · int xsinx dx = sinx -xcosx + c If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it: intu(dv)/dxdx = uv - intv(du)/dxdx , or less formally intudv=uv-intvdu I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to ...

2018年2月6日 · f' (x)=xcosx+sinx Derivation from first principles tells us that for a function f (x), f' (x)=lim_ (h->0) (f (x+h)-f (x))/h In this case, f (x)=xsinx, so we have: f ...

2016年4月11日 · intxsinxdx=-xcosx+sinx+C For this integral, we'll use integration by parts. Choose your u to be x, so that way (du)/dx=1->du=dx. That means dv=sinxdx->intdv=intsinxdx->v=-cosx. The integration by parts formula is: intudv=uv-intvdu We have u=x, du=dx, and v=-cosx. Substituting into the formula gives: intxsinxdx=-xcosx-int (-cosx)dx color (white) (XX)=-xcosx+intcosxdx color (white) (XX)=-xcosx ...

2017年11月20日 · Your regular multiplication rule We just use the general rule of multiplicative derivatives. f (x)=g (x)h (x) f' (x)=g' (x)h (x)+g (x)h' (x) The same is standard for a three part function f (x)=g (x)h (x)i (x) f' (x)=g' (x)h (x)i (x)+g (x)h' (x)i (x)+g (x)h (x)i' (x) We can get the answer simply using that statement. f (x)=xe^xsinx f' (x)= (x)'e^xsinx+x (e^x)'sinx+xe^x (sinx)' f' …

2016年10月19日 · (dy)/ (dx)= (x^sinx) (cosxlnx+sinx/x) let y=x^sinx take natural logarithms to both sides and simplify lny=lnx^sinx =>lny=sinxlnx differentiate both sides wrt x d/ (dx) (lny)=d/ (dx) (sinxlnx) using implicit differentiation on the LHS; product rule on RHS =1/y (dy)/dx=cosxlnx+sinx/x => (dy)/ (dx)=y (cosxlnx+sinx/x) substituting back for y (dy)/ (dx)= (x^sinx) (cosxlnx+sinx/x)

2018年5月10日 · Even An even function is defined as one which: f (x)=f (-x) An odd function is defined as one which: f (-x)=-f (x) We have f (x)=xsinx f (-x)=-xsin (-x) Due to the nature of sinx, sin (-x)=-sinx So, f (-x)=-x*-sinx=xsinx=f (x) f (x)=f (-x) xsinx is therefore even,

2016年7月18日 · The limit does not exist. See below. We can determine the result by pure intuition. We know that sinx alternates between -1 and 1, from negative infinity to infinity. We also know that x increases from negative infinity to infinity. What we have, then, at large values of x is a large number (x) multiplied by a number between -1 and 1 (due to sinx). This means the limit does not exist. We do ...