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differentiation of sin x + cos x.
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How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
What the differentiation and anti-differentiation formula?
There is no single formula for differentiation and anti-differentiation.The deriviative of a function y = f(x) is the limit of delta y over delta x as delta x approaches zero.OR:If f(x)=axn,f'(x)=(an)xn-1The deriviative of 2x3 would be 6x2.The anti-deriviative of a function is the reverse operation, i.e. the function is the deriviative of the anti-deriviative.Anti differentiation introduction:Anti differentiation is also called as integration process. It gives the reverse value of the differentiation equation. Anti differentiation is also called as anti derivative of the function. In this anti differentiation, f(x) is anti derivative of the function F(x). Anti differentiation is used for finding the area of the region under the certain curve. Anti differentiation symbol is denoted as ∫.General formula for anti differentiation:∫ xn dx = [xn + 1 / (n + 1)]+ c∫ k dx = k ∫ dx∫ udv = uv - ∫ v du∫ (w + y) dx = ∫ w dx + ∫ y dxanti-differentiation
How do you solve the following identity sec x - cos x equals sin x tan x?
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
Differentiation of sin x 1 plus cosx?
The differentiation of sin x plus cosx is cos (x)-sin(x).
What is the differentiation of sin x over 1 plus cosx?
The derivative is 1/(1 + cosx)
What is the differentiation of say -x sine 4x?
-x*4*cos4x + (-1)*sin4x = -4xcos(4x) - sin(4x)
What is the differentiation of 'sin 4x'?
d/dx[sin(4x)] = sin(4x) ======
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
What is forward differentiation in numerical differentiation?
( u(x+h) - u(x) )/ h
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
What is the relation between ordinary differentiation and partial differentiation?
in case of partial differentiation , suppose a z is a function of x and y so in partial differentiation of z w.r.t x all other variables except x are considered to be constant but on the contrary in differentiation process they are not considered as constant unless stated .
What does cosx divided by 1-sinx equal?
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
What is sin x times sin x?
We write sin x * sin x = sin2 x
What is the implicit differentiation of y equals sin x plus y?
y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]
