limx→−32(4x2−6x)(4x2+6x+9)2x−3 \lim_{x \to -\frac{3}{2}} \frac{(4x^2 - 6x)(4x^2 + 6x + 9)}{\sqrt{2x - \sqrt{3}}} x→−23lim2x−3(4x2−6x)(4x2+6x+9)
f(x)={(4x−1)4cot(xlog4)sin(xlog4)log(1+x2log4),if x≠0k,if x=0 f(x) = \begin{cases} \frac{(4^x - 1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} f(x)={sin(xlog4)log(1+x2log4)(4x−1)4cot(xlog4),k,if x=0if x=0
Find ek e^k ek if f(x) f(x) f(x) is continuous at x=0 x = 0 x=0.
y=sin(log(2x))+sin(log(2x))+sin(log(2x))+…∞ y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}} y=sin(log(2x))+sin(log(2x))+sin(log(2x))+…∞
Find dydx \frac{dy}{dx} dxdy for the given function:
y=tan−1(sin3(2x)−3x2sin(2x)3xsin(2x)−x3). y = \tan^{-1} \left( \frac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin(2x) - x^3} \right). y=tan−1(3xsin(2x)−x3sin3(2x)−3x2sin(2x)).
The length of the normal drawn at t=π4 t = \frac{\pi}{4} t=4π on the curve x=2(cos2t+tsin2t) x = 2(\cos 2t + t \sin 2t) x=2(cos2t+tsin2t), y=4(sin2t+tcos2t) y = 4(\sin 2t + t \cos 2t) y=4(sin2t+tcos2t) is:
If water is poured into a cylindrical tank of radius 3.5 ft at the rate of 1 cubic ft/min, then the rate at which the level of the water in the tank increases (in ft/min) is:
The function y=2x3−8x2+10x−4 y = 2x^3 - 8x^2 + 10x - 4 y=2x3−8x2+10x−4 is defined on [1,2][1,2][1,2]. If the tangent drawn at a point (a,b) (a,b) (a,b) on the graph of this function is parallel to the X-axis and a∈(1,2) a \in (1,2) a∈(1,2), then a= a = a= ?
If m m m and M M M are respectively the absolute minimum and absolute maximum values of a function f(x)=2x3+9x2+12x+1 f(x) = 2x^3 + 9x^2 + 12x + 1 f(x)=2x3+9x2+12x+1 defined on [−3,0][-3,0][−3,0], then m+M m + M m+M is:
Evaluate the integral: ∫secx3(secx+tanx)+2 dx \int \frac{\sec x}{3(\sec x + \tan x) + 2} \,dx ∫3(secx+tanx)+2secxdx
Evaluate the integral: ∫dx4+3cotx \int \frac{dx}{4 + 3\cot x} ∫4+3cotxdx
Evaluate the integral: ∫dx(x+1)x2+4 \int \frac{dx}{(x+1)\sqrt{x^2 + 4}} ∫(x+1)x2+4dx
Evaluate the integral: ∫π53π10dxsec2x+(tan2022x−1)(sec2x−1) \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{dx}{\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1)} ∫5π103πsec2x+(tan2022x−1)(sec2x−1)dx