How can I differentiate sin3x using chains rule?
Let y = sin(3x)Let u = 3xdu/dx = 3y = sin(u)dy/du = cos(u)Chain Rule:dy/dx = (dy/du)(du/dx) = 3cos(u) = 3cos(3x)
How to differentiate y=sin(ex^-5 + cos3x)?
Let u = e^(x^-5) + cos3x , y = sinu Then du/dx = (-5x^-6)e^(x^-5) - 3sin(3x) dy/du = cosu But dy/dx = dy/du * du/dx = cos(e^(x^-5) + cos3x) * (-5x^-6)e^(x^-5) - 3sin(3x)
How do I differentiate [math]\sin^3x[/math] w.r.t. [math]\cos^2x[/math]?
[math]y = \sin^3x[/math]Taking derivative…[math]\dfrac{dy}{dx} = 3\sin^2x \cos x[/math]And let's assume [math]t = \cos^2x[/math]Taking derivative…[math]\dfrac{dt}{dx} = -2\cos x \sin x[/math]We basically need to find [math]\dfrac{dy}{dt}[/math]We can do a mischief…[math]\dfrac{dy}{dt} = \dfrac{\frac{dy}{dx}}{\frac{dt}{dx}}[/math][math]\dfrac{dy}{dt} = \dfrac{3\sin^2x \cos x}{-2\cos x \sin x}[/math][math]\dfrac{dy}{dt} = \dfrac{-3}{2}\sin x[/math]Solved it :D
What is the differentiation of sin^3(x)?
*You have to apply chain rule.*basic formulae:-d/dx (x^n) = nx^n-1d/dx (sin x) = cos xSolve it as follows :-d/dx (sin^3x) = 3 sin^2x. d/dx (sin x)Therefore = 3 sin^2x.cos xThat's it. Hope it helps you. :)
What is the derivative of sin^4(x) +cos^4(x). How come it'll become -sin4x?
let, y = sin^4(x) + cos^4(x)Differentiating both side w.r.t x and applying chain ruledy/dx = 4sin^3(x)cos(x) - 4cos^3(x)sin(x)= 4sin(x)cos(x){sin^2(x) - cos^2(x)}= 2* 2sin(x)cos(x) [ - {cos^2(x) - sin^2(x)}]= - 2 sin(2x)[cos(2x)]= - 2 sin(2x)cos(2x)= - sin(4x)** Formula usedI . Sin2x = 2sinxcosx , cos2x = cos^2x - sin^2xII. Sin4x = 2 sin2x cos2xIII. d(sinx)/dx = cosxIV. d(cosx)/dx = - sinx
Logarithmic differentiation to find the derivative of the function: y=(sin(3x))^(ln(x))?
... y = (sin(3x))^(ln(x)) or ln(y) = ln(x) sin(3x) or 1/y dy/dx = (1/x) sin(3x) + 3 ln(x) cos(3x) or dy/dx = y [ (1/x) sin(3x) + 3 ln(x) cos(3x) ] or dy/dx = (sin(3x))^(ln(x)) [ (1/x) sin(3x) + 3 ln(x) cos(3x) ]
Diffrentiate the implicit functions x sin 2y - y cos 3x=5?
sin2y + 2xcos2y y' - y' cos3x + 3ysin3x = 0, (2xcos2y - cos3x) y' = - sin2y - 3ysin3x, (cos3x - 2xcos2y) y' = sin2y + 3ysin3x, y' = (sin2y + 3ysin3x)/(cos3x - 2xcos2y)
Given h(x) = 3 cos 3x - sin 5x, find g^(18)(x), the 18th derivative of g(x)?
h(x) = 3 cos 3x - sin 5x h'(x) = -3 sin 3x • 3 - cos 5x • 5 = -9 sin 3x - 5 cos 5x = -3^2 sin 3x - 5 cos 5x h"(x) = -9 cos 3x • 3 + 5 sin 5x • 5 = -27 cos 3x + 25 sin 5x = -3^3 cos 3x + 5^2 sin 5x h^(3) (x) = -3^4 sin 3x + 5^3 cos 5x h^(4) (x) = -3^5 cos 3x - 5^4 sin 5x (This has a period of 4, so h^(18) has the same pattern as h^2) ... h^(18) (x) = -3^19 cos 3x + 5^18 sin 5x
What is the second order derivative of sin3x.cos2x?
Let y= sin 3x.cos 2xy`= sin 3x.(-2 sin 2x) + cos 2x. 3 cos 3xy' =-2sin 3x.sin 2x + 3 cos 3x. Cos 2xy'' = -2{ sin 3x.(2 cos 2x)+sin 2x.3sin 3x} +3{cos 3x.(-2sin 2x)+cos 2x (-3sin 3x)}y’’= -4sin 3x .Cos 2x-6sin 2x sin 3x-6cos 3x.sin 2x -9 cos 2x.sin 3xy'' =-13sin 3x.cos 2x-6 sin 2xsin 3x-6 cos 3x.sin 2xFor more differentiation tricks and short cut [1]visit my youtube channelFootnotes[1] Differentiation - YouTube