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Derivation of trigonometric function Y = (sin3x) ^ 2 + 5cos x ^ 2 is known to find the derivative of Y!
Y=(Sin3x)^2+5Cos x^2
y'=(sin3x)'sin3x+sin3x(sin3x)'+5（cosx^2）'
=3cos3xsin3x+3cos3xsin3x-10xsinx^2
=6cos3xsin3x-10xsinx^2
=3sin6x-10xsinx^2
Why is trigonometric function still trigonometric function after derivation Is the derivative of an ordinary trigonometric function still a trigonometric function?
affirmative.
The derivative of a function reflects the change law of the tangent slope of the function. You can draw a diagram and try. No matter which point, the tangent slope of a trigonometric function is expressed by a trigonometric function (just another trigonometric function), so its derivative is still a trigonometric function
If the domain of ax ^ 2-ax + 1 / A under the function f (x) = root sign is all real numbers, then the value range of real number a
That is, ax ^ 2-ax + 1 / a > = 0 is always true
Because a ≠ 0
So this is a quadratic function
Constant ≥ 0
So open up
a>0
And △
The function f (x) = log (2) [(AX-1) / (x ^ 2-x + 2) + 2)] is always meaningful when x belongs to [1,3], then the value range of real number a
x∈[1,3],(ax-1)/(x ²- x+2)+2>0x ²- X + 2 is always greater than 0 on R; ∴ax-1+2(x ²- x+2)>0; a>-2x+2-3/x; 2x+3/x≥2√6; If and only if 2x = 3 / x, there is a minimum value of 2 √ 6; That is, x = √ 6 / 2 ∈ [1,3]; Conform to ‡ - 2x-3 / X ≤ - 2 √ 6; ∴a>-2√6+2...
Using Lim SiNx / x = 1 or equivalent infinitesimal to find the limit LIM (COS) α x-cos β x) / x ^ 2 x tends to 0
Use trigonometric function formula: cos α x-cos β x=-2sin(( α x+ β x)/2)sin(( α x- β x) / 2) original formula = lim - 2Sin（（ α x+ β x)/2)sin(( α x- β x)/2)/x ² Equivalent Infinitesimal Substitution: sin（（ α x+ β x)/2) ( α x+ β x)/2sin(( α x- β x)/2) ( α x- β x) / 2 original formula = - 2
Find the limit LIM ((x → X / 2) cosx) / (COS (x / 2) - sin (x / 2))
X - > π / 2, right? Cosx = sin (π / 2-x) because π / 2-x - > 0, sin (π / 2-x) ~ (π / 2-x) pair denominator cos (x / 2) - sin (x / 2) = √ 2 [(√ 2) / 2) cos (x / 2) - (√ 2) / 2) sin (x / 2)] = √ 2 [sin (π / 4) cos (x / 2) - cos (π / 4) sin (x / 2)] = √ 2 sin (π / 4-x / 2) = √ 2 sin
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