All Answers 358525 Answers

CPHILL REPORT HER BANN HER No rude comments young lady don't be calling people b!tches unless you wanna be called one thats what we say to guest you're not any diffrent

Derivative:f(x)=(cosx)^(cosx)

Find the derivative of the following via implicit differentiation: d/dx(f(x)) = d/dx(cos^(cos(x))(x)) The derivative of f(x) is f'(x): f'(x) = d/dx(cos^(cos(x))(x)) Express cos^(cos(x))(x) as a power of e: cos^(cos(x))(x) = e^(log(cos^(cos(x))(x))) = e^(cos(x) log(cos(x))): f'(x) = d/dx(e^(cos(x) log(cos(x)))) Using the chain rule, d/dx(e^(cos(x) log(cos(x)))) = ( de^u)/( du) ( du)/( dx), where u = cos(x) log(cos(x)) and ( d)/( du)(e^u) = e^u: f'(x) = d/dx(cos(x) log(cos(x))) e^(cos(x) log(cos(x))) Express e^(cos(x) log(cos(x))) as a power of cos(x): e^(cos(x) log(cos(x))) = e^(log(cos^(cos(x))(x))) = cos^(cos(x))(x): f'(x) = cos(x)^(cos(x)) d/dx(cos(x) log(cos(x))) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = log(cos(x)): f'(x) = log(cos(x)) d/dx(cos(x))+cos(x) d/dx(log(cos(x))) cos^(cos(x))(x) The derivative of cos(x) is -sin(x): f'(x) = cos^(cos(x))(x) (cos(x) (d/dx(log(cos(x))))+-sin(x) log(cos(x))) Using the chain rule, d/dx(log(cos(x))) = ( dlog(u))/( du) ( du)/( dx), where u = cos(x) and ( d)/( du)(log(u)) = 1/u: f'(x) = cos^(cos(x))(x) (-(log(cos(x)) sin(x))+d/dx(cos(x)) sec(x) cos(x)) Simplify the expression: f'(x) = cos^(cos(x))(x) (d/dx(cos(x))-log(cos(x)) sin(x)) The derivative of cos(x) is -sin(x): f'(x) = cos^(cos(x))(x) (-(log(cos(x)) sin(x))+-sin(x)) Expand the left hand side: Answer: | | f'(x) = cos^(cos(x))(x) (-sin(x)-log(cos(x)) sin(x))

Derivative: 3y^3-4x^2y+xy=-5

Find the derivative of the following via implicit differentiation: d/dx(x y-4 x^2 y+3 y^3) = d/dx(-5) Differentiate the sum term by term and factor out constants: d/dx(x y)-4 d/dx(x^2 y)+3 d/dx(y^3) = d/dx(-5) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = y: -4 (d/dx(x^2 y))+3 (d/dx(y^3))+x d/dx(y)+d/dx(x) y = d/dx(-5) Simplify the expression: x (d/dx(y))-4 (d/dx(x^2 y))+3 (d/dx(y^3))+(d/dx(x)) y = d/dx(-5) The derivative of y is y'(x): -4 (d/dx(x^2 y))+3 (d/dx(y^3))+y'(x) x+(d/dx(x)) y = d/dx(-5) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^2 and v = y: 3 (d/dx(y^3))-4 x^2 d/dx(y)+d/dx(x^2) y+(d/dx(x)) y+x y'(x) = d/dx(-5) The derivative of y is y'(x): 3 (d/dx(y^3))+(d/dx(x)) y-4 (y'(x) x^2+(d/dx(x^2)) y)+x y'(x) = d/dx(-5) Using the chain rule, d/dx(y^3) = ( du^3)/( du) ( du)/( dx), where u = y and ( d)/( du)(u^3) = 3 u^2: 3 3 d/dx(y) y^2+(d/dx(x)) y+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) Simplify the expression: (d/dx(x)) y+9 (d/dx(y)) y^2+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) The derivative of y is y'(x): (d/dx(x)) y+y'(x) 9 y^2+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) The derivative of x is 1: 1 y+x y'(x)+9 y^2 y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x: y+x y'(x)+9 y^2 y'(x)-4 (2 x y+x^2 y'(x)) = d/dx(-5) The derivative of -5 is zero: y+x y'(x)+9 y^2 y'(x)-4 (2 x y+x^2 y'(x)) = 0 Expand the left hand side: y-8 x y+x y'(x)-4 x^2 y'(x)+9 y^2 y'(x) = 0 Subtract y-8 x y from both sides: x y'(x)-4 x^2 y'(x)+9 y^2 y'(x) = -y+8 x y Collect the left hand side in terms of y'(x): (x-4 x^2+9 y^2) y'(x) = -y+8 x y Divide both sides by -4 x^2+x+9 y^2: Answer: | | y'(x) = (-y+8 x y)/(x-4 x^2+9 y^2)

y^3+sinhxy^2=3/2

Find the derivative of the following via implicit differentiation: d/dx(sinh^2(x y)+y^3) = d/dx(3/2) Differentiate the sum term by term: d/dx(sinh^2(x y))+d/dx(y^3) = d/dx(3/2) Using the chain rule, d/dx(sinh^2(x y)) = ( du^2)/( du) ( du)/( dx), where u = sinh(x y) and ( d)/( du)(u^2) = 2 u: d/dx(y^3)+2 d/dx(sinh(x y)) sinh(x y) = d/dx(3/2) Using the chain rule, d/dx(sinh(x y)) = ( dsinh(u))/( du) ( du)/( dx), where u = x y and ( d)/( du)(sinh(u)) = cosh(u): d/dx(y^3)+cosh(x y) d/dx(x y) 2 sinh(x y) = d/dx(3/2) Using the chain rule, d/dx(y^3) = ( du^3)/( du) ( du)/( dx), where u = y and ( d)/( du)(u^3) = 3 u^2: 2 cosh(x y) (d/dx(x y)) sinh(x y)+3 d/dx(y) y^2 = d/dx(3/2) The derivative of y is y'(x): 2 cosh(x y) (d/dx(x y)) sinh(x y)+y'(x) 3 y^2 = d/dx(3/2) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = y: x d/dx(y)+d/dx(x) y 2 cosh(x y) sinh(x y)+3 y^2 y'(x) = d/dx(3/2) The derivative of y is y'(x): 2 cosh(x y) sinh(x y) (y'(x) x+(d/dx(x)) y)+3 y^2 y'(x) = d/dx(3/2) The derivative of x is 1: 3 y^2 y'(x)+2 cosh(x y) sinh(x y) (1 y+x y'(x)) = d/dx(3/2) The derivative of 3/2 is zero: 3 y^2 y'(x)+2 cosh(x y) sinh(x y) (y+x y'(x)) = 0 Expand the left hand side: 2 cosh(x y) sinh(x y) y+2 x cosh(x y) sinh(x y) y'(x)+3 y^2 y'(x) = 0 Subtract 2 y sinh(x y) cosh(x y) from both sides: 2 x cosh(x y) sinh(x y) y'(x)+3 y^2 y'(x) = -2 cosh(x y) sinh(x y) y Collect the left hand side in terms of y'(x): (2 x cosh(x y) sinh(x y)+3 y^2) y'(x) = -2 cosh(x y) sinh(x y) y Divide both sides by 2 x sinh(x y) cosh(x y)+3 y^2: Answer: | | y'(x) = -(2 cosh(x y) sinh(x y) y)/(2 x cosh(x y) sinh(x y)+3 y^2)

Let y=sin(n*sin^-1x), n>0.Show that: (1 - x^2)*y" - xy' + n^2y=0

n>0,   y = sin(n sin^(-1)(x))

P.S. YOU HAVE TO RE-SUBMIT THE LAST TWO IN HAND-WRITING. DO NOT USE A MATH PACKAGE SUCH AS LA TEX.....ETC., BECAUSE WE CANNOT COPY AND PASTE THEM!.

The ratio10:17  means that there are 27 "parts"   to his collection and the Blu-rays  comprise 10/27 of the total.......so....

10/27 * 837   =  310 Blu-rays

Good bye I guess... ｡゜(｀Д´)゜｡

I need some hugs... Pls... ༼ つ ಥ_ಥ ༽つ

-2x+7>0     subtract 7 from each side

-2x  > -7    divide both sides by -2   and reverse the inequality sign

x < 7/2

4.53*10⁹ I believe... 

What's up?

35/2 = 17.5

x = 17.5

4.8/6 = 0.8

Yeah. She did, according to her mom. I texted her hi, and her mom said that her computer was confiscated, and she was in trouble.

yea, she left :/

how do you work out 3^2010

3^2010=1.0321 X 10^959

wat happen to her

lim x-->9 (x sqrt(3) - 3) / (x - 9)

(two-sided limit does not exist)

lim_(x->9^-) (x sqrt(3)-3)/(x-9) = -infinity (Limit from the left)

lim_(x->9^+) (x sqrt(3)-3)/(x-9) = infinity (limit from the right)

lim x-->∞ (3 + 2/x) * cos(1/x)

Find the following limit:

lim_(x->infinity) cos(1/x) (3+2/x)

Applying the product rule, write lim_(x->infinity) (3+2/x) cos(1/x) as (lim_(x->infinity) (3+2/x)) (lim_(x->infinity) cos(1/x)):

lim_(x->infinity) (3+2/x) lim_(x->infinity) cos(1/x)

Using the fact that cosine is a continuous function, write lim_(x->infinity) cos(1/x) as cos(lim_(x->infinity) 1/x):

lim_(x->infinity) (3+2/x) cos(lim_(x->infinity) 1/x)

Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so

lim_(x->infinity) 1/x = 0:

cos(0) lim_(x->infinity) (3+2/x)

cos(0) = 1:

lim_(x->infinity) (3+2/x)

3+2/x = lim_(x->infinity) 3+2 (lim_(x->infinity) 1/x):

lim_(x->infinity) 3+2 lim_(x->infinity) 1/x

Let epsilon>0. Then for all x > 1/epsilon, 1/x<1/(1/epsilon) = epsilon, so

lim_(x->infinity) 1/x = 0:

lim_(x->infinity) 3+2×0

Since 3 is constant, lim_(x->infinity) 3 = 3:

3+2×0

3+2 0 = 3:

Answer: |=3

lim x-->0 ((1 -  cos(ax)) / x^2)

Find the following limit: lim_(x->0) (1-cos(a x))/x^2 Applying l'Hôpital's rule, we get that lim_(x->0) (1-cos(a x))/x^2 = lim_(x->0) ( d/( dx)(1-cos(a x)))/( d/( dx) x^2) = lim_(x->0) (a sin(a x))/(2 x) lim_(x->0) (a sin(a x))/(2 x) (a lim_(x->0) (sin(a x))/x)/(2) Applying l'Hôpital's rule, we get that lim_(x->0) (sin(a x))/x | = | lim_(x->0) ( d/( dx) sin(a x))/(( dx)/( dx)) | = | lim_(x->0) (a cos(a x))/1 | = | lim_(x->0) a cos(a x) (a lim_(x->0) a cos(a x))/2 lim_(x->0) a cos(a x) = a cos(0 a) = a: (a a)/2 1/2 a a = a^2/2: Answer: | | a^2/2

lim x-->∞ (x^1/3 - 5x + 3) / (2x + x^2/3 - 4)

Find the following limit: lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) = (3+x^(1/3)-5 x)/(x^2/3+2 x-4): lim_(x->infinity) (3+x^(1/3)-5 x)/(x^2/3+2 x-4) The leading term in the denominator of (3+x^(1/3)-5 x)/(x^2/3+2 x-4) is x^2. Divide the numerator and denominator by this: lim_(x->infinity) (3/x^2+1/x^(5/3)-5/x)/(1/3+2/x-4/x^2) The expressions 3/x^2, 1/x^(5/3), -5/x, -4/x^2 and 2/x all tend to zero as x approaches infinity: 0/(1/3) 0/(1/3) = 0: Answer: | | 0

lim x-->∞ (sqrt(x^2 + x)) - sqrt(x^2 - cx), c=positive constant

You didn't specify the limit:lim_(x->infinity) (sqrt(x^2+x)-sqrt(x^2-c x)) = (1+c)/2

2888267505509060205.86/9994299999999999999999999999999999999999998605800000000000000000000000000000000000000000000129.41 = 0

f(4)=3(4)-7

f(4)=12-7

f(4)=5

2^128 = 340282366920938463463374607431768211456

\(\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

3^2010 = infinity

You are multiply 3 by 3 by 3 by 3 by 3 ............................

You get me right?

It is infinity because it almost never ends.

Wait she's leaving?? 

 um... wow thats mature guys

7-8=-1-(-5)=-1+5=4-1=3

The prime factorization of 98 is as follows:

                            98

                           /   \

                         2    49

                               /   \

                             7     7 

Prime factorization in an equation is: 2*7*7

(1/4)^4 = 1/256 = y