Let y = sec (tan (sqrt(x))) rArr (dy)/(dx)= (d)/(dx) sec (tan (sqrt(x))) = sec(tansqrt(x)). tan (tansqrt(x)).(d)/(dx)(tansqrt(x)) = sec(tansqrt(x)). tan (tansqrt(x)).sec^(2) sqrt(x) .(d)/(dx)sqrt(x) = (sec(tansqrt(x)). tan (tansqrt(x))....
Differentiate the following functions with respect tox:tan−1(secx+tanx),−π2<x<π2 View Solution Differentiate the following functions with respect tox:tan−1(secx+tanx),−π2<x<π2 View Solution Differentiate(tanx−13secx)with respect to 'x'. ...
1−sec(x)tan(x) Differentiation: Properties of Differentiation: 1.d(uv)=vdu−udvv2 2.d(uv)=udv+vdu. Formulas: 1.d(tanx)=sec2xdx. 2.d(secx)=secxtanx dx. Answer and Explanation: Given y=1−secxtanx Differentiating with respect to x...
Use the Quotient Rule to differentiate the functionf(x)=tan(x)−1sec(x). Question: Use the Quotient Rule to differentiate the functionf(x)=tan(x)−1sec(x). Quotient Rule: The quotient rule is one of the rules used while performing differentiation. ...
1.基本初等函数的导数 常见的基本初等函数的导数如下: - cos(x) 的导数为 -sin(x) - sin(x) 的导数为 cos(x) - tan(x) 的导数为 sec^2(x) - log(x) 的导数为 1/x - e^x 的导数为 e^x - ax 的导数为 a 2.复合函数的导数 复合函数的导数遵循链式法则,即外函数的导数乘以内函数的导数。
Differentiate the following: y = (2x+1)^2 \sqrt{x^2 + 1}. Differentiate the following. x^2 - 4 x y + 3 y sin x = 17 Differentiate the following y=4^{6x} . Differentiate the following. y = x^2 sin x tan x Differentiate the following: y = 3x^3 - 5x^{-2} + 9x ...
k(1+4x)^9 4 ×10(1+4x)9 or better (a) (1+4x)^(10)(their-sinx))+ cosx(their(4*10)*(1+4x)^9) (1+4x)^(10)(-sinx)+cosx(4*10^(π/4))*(1+4x)^9 d/(dx)(e^(4x-5))=4e^(4x-5)soi d/(dx)(tanx)=sec^2xsoi clearly applies correct form of quotient rule (...
1.[tan(2x+1)]'=sec²(2x+1)*(2x+1)'=2sec²(2x+1);2.因为2sec²(2x+1)恒>0,所以tan(2x+1)没有驻点.(驻点就是导数为0的点)3.对于一个很小的p,找出函数tan(2x+1)当x从1增加到1+p时的近似该变量.其实就是求微分:x=1,dx=p时,dtan(2x+1)=2sec²(2x+1)dx=2psec²3...
A) {eq}y = \frac{1}{2} \tan^{-1} x + \frac{x}{2(x^2 + 1)} {/eq} B) {eq}f(x) = \frac{\arctan x}{1 + x^2} {/eq} Derivatives of Inverse Trigonometric Functions: Here are the derivatives of some common inverse trigonometric functions: {eq}\begi...
Putting x=tanθ, we get y=tan−1(secθ+1tanθ)=tan−1(1+cosθsinθ) =tan−1{2cos2(θ/2)2sin(θ/2)cos(θ/2)}=tan−1{cot.θ2} =tan−1{tan(π2−θ2)}=(π2−θ2)=π2−12tan−1x. ∴dydx=12(1+x2). Hence, ddx{tan−1(√1+x2+1x)}=−12...