y' = dy/dx = y的无穷小增量除以x的无穷小增量,所以dy/dx又称为“微商”. dy = 就是对y的微分 dx = 就是对x的微分 对函数f(x)的微分: df = [df/dx]dx = f ' dx df/dx 就是对原函数 f(x) 求导得到的导函数(derivative)结果一 题目 不定积分公式,为什么∫f(sinx)cosxdx=∫f(sinx)d(...
Answer to: Find the derivative of f(x)=xcosx+sinx By signing up, you'll get thousands of step-by-step solutions to your homework questions. You can...
『例子一』 y=sinx, y'=cosx 『例子二』 y=e^x, y'=e^x 『例子三』 y=cosx, y'=-sinx 👉回答 f(x)= sinx/(1- cosx)利用 (u/v)' = (vu'-uv')/v^2 f'(x)=[(1- cosx).(sinx)' - sinx.(1- cosx)']/(1- cosx)^2 =[(1- cosx).(cosx) - sinx.(sin...
∵ ∫f(x)dx = xsinx + c [Given， 已知]∴ f(x) = sinx + xcosx [Derivative， 求导]∴ ∫xf'(x)dx = ∫xdf(x) [Completing differentiation，凑微分]= xf(x) - ∫f(x)dx + c [Integration by parts，分部积分]= x(sinx + xcosx) - xsinx + c [Substit...
∵∫f(x)dx=xsinx+c[Given,已知] ∴f(x)=sinx+xcosx[Derivative,求导] ∴∫xf'(x)dx=∫xdf(x)[Completingdifferentiation,凑微分] =xf(x)-∫f(x)dx+c[Integrationbyparts,分部积分] =x(sinx+xcosx)-xsinx+c[Substitution,代入] =x²cosx+c[Simplification,化简] 相关...
又:y' = dy/dx = y的无穷小增量除以x的无穷小增量,所以dy/dx又称为“微商”.dy = 就是对y的微分dx = 就是对x的微分对函数f(x)的微分:df = [df/dx]dx = f ' dxdf/dx 就是对原函数 f(x) 求导得到的导函数(derivative) 解析看不懂?免费查看同类题视频解析查看解答...
∵∫f(x)dx = xsinx + c [Given,已知]∴ f(x) = sinx + xcosx [Derivative,求导]∴∫xf'(x)dx = ∫xdf(x) [Completing differentiation,凑微分]= xf(x) - ∫f(x)dx + c [Integration by parts,分部积分]= x(sinx + xcosx) - xsinx + c [Substitution,代入]= x²cosx + c [Simpl...
导数（Derivative），也叫导函数值。又名微商，是微积分中的重要基础概念。当函数y=f（x）的自变量x在一点x0上产生一个增量Δx时，函数输出值的增量Δy与自变量增量Δx的比值在Δx趋于0时的极限a如果存在，a即为在x0处的导数，记作f'（x0）或df（x0）/dx。常用导数公式：1、y=c(c为常数)...
∵∫f(x)dx = xsinx + c [Given,已知]∴ f(x) = sinx + xcosx [Derivative,求导]∴∫xf'(x)dx = ∫xdf(x) [Completing differentiation,凑微分]= xf(x) - ∫f(x)dx + c [Integration by parts,分部积分]= x(sinx + xcosx) - xsinx + c [Substitution,代入]= x²cosx + c [Simpli...
To solve the problem, we need to find the value of the n-th derivative of the function f(x) at x=0, where f(x)=∣∣∣∣∣xnsinxcosxn!sin(nπ/2)cos(nπ/2)aa2a3∣∣∣∣∣ for n=2m+1. Step 1: Write the determinantThe function f(x) can be expressed as a determinant of...