Info From the Get-go (Extra information that is important)
This function is the reflective line for inverse functions.
What is y=x?
This inverse trigonometric function has the derivative of 1/(x2+1).
What is arctan(x)?
The differentiation of h(x)=tan-1(sinx) is this.
What is cosx/(1+sin2x)?
h(x)=tan-1(sinx)
h'(x)=1/1+(sinx)2 *(cosx)
h'(x)=cosx/(1+sin2x)
f(x) and g(x) are inverses. This answer is the solution to f'(5) when f(5)=4 and g'(x)=x3.
What is 1/96?
f(x) and g(x) are inverses; g(x)=2x3
g(x)=2x3
g'(x)=6x2
f'(x)=1/g'(f(x))
f'(6)=1/g'(f(5))
f'(6)=1/g'(4)
f'(6)=1/(6(4)2)
f'(6)=1/96
What is g'(x)=1/f'(g(x))
The derivative of arcsin(x) is equivalent to this.
What is 1/sqrt(1-x2)?
This solution is the result of Differentiating y=arcsinx-x*sqrt(1-x^2).
What is y=2sqrt(1-x^2)?
y=arcsinx-x*sqrt(1-x^2)
y'=1/sqrt(1-x^2)+(x)(-2x)(1/2)(1-x^2)^-1/2+sqrt(1-x^2)
y'=1/sqrt(1-x^2)-(x^2)/sqrt(1-x^2)+(1-x^2)/sqrt(1/x^2)
y'=(2-2x^2)/sqrt(1-x^2)
y'=2(1-x^2)/sqrt(1-x^2)
y'=2sqrt(1-x^2)
k(x) and t(x) are inverses. The solution of k'(9) with k(9)=12 and t(x)=12x-3x2.
What is -1/60?
k(x) and t(x) are inverses; t(x)=12x-3x2; k(9)=12
t(x)=12x-3x2
t'(x)=12-6x
k'(x)=1/t'(k(x))
k'(9)=1/t'(k(9))
k'(9)=1/t'(12)
k'(9)=1/(12-6(12))
k'(9)=1/(-60)
k'(9)=-1/60
This is the Domain restriction of the functions arcsinx and arccosx.
What is -1<or=x<or=1?
The original function of this derivative is written as arccot(t).
What is -1/(x2+1)?
This solution is the result of differentiating the function y=arccos(3x2).
What is -6x/sqrt(1-9x4)?
y=arccos(3x2)
y'=-1/sqrt(1-(3x2)2) * (6x)
y'=-6x/sqrt(1-9x4)
q(x) and h(x) are inverses. This answer is the solution of h'(6) when q(x)=x3-4x2 and q(5)=6.
What is 1/35?
q(x) and h(x) are inverses; q(x)=x3-4x2; q(5)=6
If h(x) and q(x) are inverses and q(5)=6, then that means h(6)=5
q(x)=x3-4x2
q'(x)=3x2-8x
h'(x)=1/q'(h(x))
h'(6)=1/q'(h(6))
h'(6)=1/q'(5)
h'(6)=1/(3(5)2-8(5))
h'(6)=1/(75-40)
h'(6)=1/35
This is the Domain restriction of the functions arctanx and arccotx.
What is -infinity<or=x<or=infinity?
The derivative of this function is equivalent to 1/(|w|sqrt(w2-1)).
What is arcsec(w)?
The derivative of the function h(x)=5x2arccos(x)-2x is this.
What is 10xarccos(x)-5x2/sqrt(1-x2) -2
h(x)=5x2arccos(x)-2x
h'(x)=2(5x)2-1arccos(x)+5x2(-1/sqrt(1-x2) -2x1-1
h'(x)=10xarccos(x)-5x2/sqrt(1-x2) -2
f(c) and r(c) are inverses. This value is the solution of f'(3) when r(c)=5x2-sqrt(x) and r(4)=3.
What is 4/31?
f(c) and r(c) are inverses; r(c)=x2-sqrt(x); r(4)=3
Because f(c) and r(c) are inverses, and r(4)=3, f(3)=4.
r(c)=x2-sqrt(x)
r'(c)=2x-1/2sqrt(x)
f'(x)=1/r'(f(x))
f'(3)=1/r'(f(3))
f'(3)=1/r'(4)
f'(3)=1/(2(4)-1/2sqrt(4))
f'(3)=1/(8-1/4)
f'(3)=1/(31/4)
f'(3)=1*4/((31*4)/4)
f'(3)=4/31
This is the Domain restriction of the functions arcsecx and arccscx.
What is |x|>or=1?
These functions are the derivatives of the inverses of cos(t) and csc(x). (Please answer them in the order of which they are ordered).
What is the function -1/sqrt(1-t2) and -1/(|w|sqrt(w2-1))?
This solution is the result of the derivative of arcsec(e2x).
What is 2/sqrt(e4x-1)?
arcsec(e2x)
arcsec(e2x)d/dx
2e2x/(e2xsqrt((e2x)2-1)
2/sqrt(e4x-1)
*usually, there will be a |x| on the denominator, however, since ex will always be a positive output, it isn't needed, so it is factored out*
j(t) and m(t) are inverses. This solution is the direct result of j'(17) when m(15)=17 and m(t)=4x3-12x2+14x-92.
*calculator allowed*
What is 1/3074?
j(x) and m(x) are inverses; m(x)=4x3-12x2+14x-92; m(15)=17
If j(x) and m(x) are inverses and m(15)=17, then that means j(17)=15
m(x)=4x3-12x2+14x-92
m'(x)=12x2-24x+14
j'(x)=1/m'(j(x))
j'(17)=1/m'(j(17))
j'(17)=1/m'(15)
j'(17)=1/(12(15)2-24(15)+14)
j'(17)=1/(2700-360+14)
j'(17)=1/3074