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By the chain rule, the derivative of sin(x1/2) will be the derivative of x1/2 multiplied by the derivative of the enclosing sine function. Thus,
y = sin(x1/2)
y' = (1/2)*(x-1/2)*cos(x1/2)
For further reading, you might want to consult your calculus book on the chain rule. Here is a site that (kind of) explains the chain rule, though it does have good examples: http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html
For step-by-step derivatives of functions, try Calc 101: http://calc101.com/webMathematica/derivatives.jsp
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How do you differentiate cosine squared of x?
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!
How does one differentiate between the sine rule and the sine ratio?
The sine rule(also known as the "law of sines") is: a/sin A = b/sin B = c/sin C where the uppercase letters represent angles of a triangle and the lowercase letters represent the sides opposite the angles (side "a" is opposite angle "A", and so on.) Sine Ratio(for angles of right triangles): Sine of an angle = side opposite the angle/hypotenuse written as sin=opp/hyp.
What is the derivative of 2cosx times the inverse of sinx?
d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
What is sin of -pi?
sin(-pi) = sin(-180) = 0 So the answer is 0
What do you call a less serious sin?
According to the Bible, we cannot differentiate sin as minor or major. A minor sin is also a "Sin".
Differentiate sin x with respect to x?
cos x
How do you differentiate sin sin x?
To differentiate y=sin(sin(x)) you need to use the chain rule. A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside". First, you take the derivative of the outside. The derivative of sin is cos. Then, you keep the inside, so you keep sin(x). Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx. In the end, you get y'=cos(sin(x))cos(x))
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
In God's eyes are all sin equal?
The question is asked properly: Is all SIN equal (not sins). We are the ones that differentiate it, break it into "manageable" pieces. Jesus took away the SIN of the world - John 1:29.
How do you differentiate x sinx with respect to x?
You should apply the chain rule d/dx(x.sin x) = x * d/dx(sin x) + sin x * d/dx(x) = x * cos x + sin x * 1 = x.cos x + sin x
Differentiate y equals xsinx plus cosx?
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(x) ]
How does one differentiate between the sine rule and the cosine rule?
The sine rule is a comparison of ratios: (sin A)/a = (sin B)/b = (sin C)/c. The cosine rule looks similar to the theorem of Pythagoras: c2 = a2 + b2 - 2ab cos C.
How do you differentiate x sinx with respect to y?
If you actually mean "... with respect to x", and that y is equal to this function of x, then the answer is:y = x sin(x)∴ dy/dx = sin(x) + x cos(x)
How do you differentiate cosine square root of x?
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
Can you differentiate the following function fx equals 9 times x to the 2nd power times sinxtanx?
f(x)=9x2(sin x * tan x)f'(x)= 18x(sin x * tan x) + 9x2(cos x * tan x + sec2x * sin x)there might be some identities that allow that to be simplified to look prettier
How do you differentiate cosine squared of x?
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!
