A. cos(x) B. -cos(x) C. sin(x) D. -sin(x) 相关知识点: 试题来源: 解析 A。根据求导公式,sin(x) 的导数是 cos(x)。选项 B 是 cos(x) 的相反数。选项 C 是原函数。选项 D 是 -sin(x),不符合求导结果。反馈 收藏
百度试题 结果1 题目The derivative of y = sin(x) is ___. A. cos(x) B. -cos(x) C. sin(x) D. -sin(x) 相关知识点: 试题来源: 解析 A。根据求导公式,sin(x)的导数是 cos(x)。选项 B、C、D 都不符合。反馈 收藏
y=sin2x的导数是2cos2x。拓展知识：1、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...
y=sin^(-1)((3x+4sqrt(1-x^(2)))/(5)) "Let "x=sintheta,thetain[(-pi)/(2),(pi)/(2)],y=sin^(-1)((3sintheta+4costheta)/(5)) :.y=sin^(-1)(sin(theta+alpha))"where "tanalpha=(4)/(3) Now, (pi)/(4)ltalphalt(pi)/(3)," when "x=(sqrt(3))/(2),theta=(...
Solution Share Step 1 Given y=1+sin(X)cos(X) Derivative with respect X on both sides dydx=ddx(1+sin(X)cos(X)) Explanation: The above derivative is in the form ofView the full answer Step 2 Unlock Step 3 Unlock Answer UnlockPrevious question Next questionNot...
Find the derivative. y=tan(arcsinx) Derivative of Composite Function: If we have a composition of trigonometric and inverse trigonometric functions (sayT(I(x))), then its derivative is evaluated by applying the chain rule as shown below. ...
Find the derivative: a. sin(x^2y^2)=x b. sin(x)+cos(2y)=1 Find the derivative. 8) y = {1} / {11 x^2} + {1} / {3 x} A) {2} / {11 x^3} + {1} / {3 x^2} B) - {1} / {11 x^3} + {1} / {3 x^2} C) - {2} / {11 x^3} - {1} / {3 x...
3) {eq}\csc x = \dfrac{1}{\sin x}, {/eq} among others. Answer and Explanation:1 Let us find the derivative of {eq}y = \sec(x) {/eq}. Let's redefine the function as a function of cosine. Thus we have {eq}y = \dfrac{1}{\cos x}... ...
Derivative of y = "sec"^(-1)(1/(2x^2 - 1)) is 03:16 If f(x) = (x - 1)/(4) + ((x - 1)^(2))/(12) + ((x -1)^(5))/(20) + ((x -... 03:26 If f(x)=(cosx+isinx)(cos3x+isin3x)...(cos(2n-1)x+isin(2n-1)x) then f"(... 03:01 If y=f((3x...
function), y = log_a(x) (logarithmic function), y = sin(x) (sine function), y = cos(x) (cosine function), y = tan(x) (tangent function), and y = cot(x) (cosecant function). Each formula establishes the relationship between the function y and its derivative y' for...