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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
