# Calculus

Calculus

We have to find the value of lim x--> -1 [(2x^2-x-3)/(x+1)] If we substitute x = -1 in (2x^2-x-3)/(x+1) we get 0 / 0. So L'Hopital's theorem can be applied. We now have: lim x--> -1 [(4x...

Latest answer posted February 11, 2011 12:48 am UTC

Calculus

The function given is f(x) = 3*x^6 + 3*x^(-1) - cosx. We need the definite integral of f(x) between the values of x = 0 and x = 1. F(x) = Int [ f(x)] => Int [ 3*x^6 + 3*x^(-1) - cos x dx] =>...

Latest answer posted April 14, 2011 11:59 pm UTC

Calculus

We have to find the integral of f'(x)=11e^x/(11+e^x) f'(x)=11e^x/(11+e^x) let 11 + e^x = y e^x dx = dy Int [ 11e^x/(11+e^x) dx] => Int [ 11dy/y] => 11*ln |y| + C substitute y = 11 + e^x =>...

Latest answer posted April 15, 2011 2:15 am UTC

Calculus

The gradient vector is a vector whose first component is the derivative of f(x, y) with respect to x and the second is the derivative of f(x, y) with respect to y. Here f(x, y) = x^3*y^2 - 2x...

Latest answer posted April 15, 2011 3:19 am UTC

Calculus

The antiderivative of f(x)=x*e^8x can be found using integration by parts. Int [u dv ]= u*v - Int[ v du] Let u = x => du = 1 dv = e^8x => v = e^8x/8 Int [ x*e^8x ] = (x*e^8x)/8 - Int...

Latest answer posted April 13, 2011 9:37 pm UTC

Calculus

If you mean that you required the integral of y = (e^(13x)+sin x)/10, the result is. y = (e^(13x)+sin x)/10 Int [ y dx] = Int [ (e^(13x)+sin x)/10 dx] => (1/10)* Int [ (e^(13x)+sin x) dx] =>...

Latest answer posted February 25, 2011 4:20 pm UTC

Calculus

We have to find: lim x-->2 [(x^2-4)/(x^2+x-6)] substituting x= 2 gives the indeterminate form 0/0. So we can substitute the numerator and the denominator with their derivatives using l'Hopital's...

Latest answer posted March 8, 2011 2:01 am UTC

Calculus

I assume fyx means differentiation of f with respect to y followed by differentiation with respect to x, and vice versa for fyx. fxy = d[d(x^3+8xy)/dx]/dy => fxy = d(3x^2 + 8y)/dy => fxy = 0...

Latest answer posted February 25, 2011 6:04 pm UTC

Calculus

int_0^1 x^n/(n+1)=x^(n+1)/(n+1)^2|_0^1=1/(n+1)^2 So your integral is equal to 1/(n+1)^2 which is less than 1/(n+1) because n is a natural number. I'm sorry because the above formula is hard...

Latest answer posted February 4, 2013 9:57 pm UTC

Calculus

We have to find the antiderivative of y=1/(x-1)(x+4) First let's express y as partial fractions. y = A / (x - 1) + B/(x + 4) => y=1/(x-1)(x+4) = A/(x - 1) + B/(x + 4) => Ax + 4A + Bx - B = 1...

Latest answer posted May 11, 2011 6:04 am UTC

Calculus

We have to find the derivative of y = (3x^3+3)/(3x^3-3) We use the quotient rule which state that the value of [f(x) / g(x)]' = [f'(x)*g(x) - f(x)*g'(x)]/ [g(x)]^2 Substituting the functions we...

Latest answer posted February 18, 2011 7:19 pm UTC

Calculus

The only method to find the original function if you know the integrand is by the use of integration. This is because differentiation and integration are inverse functions. Though the process of...

Latest answer posted April 14, 2011 6:21 am UTC

Calculus

We have to find the integral of f(x) = 3*sin x - 4*(tan x)^2 f(x) = 3*sin x - 4*(tan x)^2 => 3*sin x - 4*[(sec x)^2 - 1] Int [ f(x) dx] = Int [3*sin x - 4*[(sec x)^2 - 1] => 3*Int [ sin x dx]...

Latest answer posted March 11, 2011 4:27 pm UTC

Calculus

We have to find the definite integral for y=x+1/x between x=1 and x=3. First let's find the indefinite integral of y Int [ y dx] = Int [ x + 1/x dx] => x^2 / 2 + ln |x| +C For x = 3 this is...

Latest answer posted May 3, 2011 1:13 pm UTC

Calculus

The chain rule of differentiation gives the derivative of f(g(x)) as f'(g(x))*g'(x) For f(x) = (x^3 + 4)^4 we can take u(x) = x^3 + 4 f(x) = (u(x))^4 f'(x) = 4*(u(x))^3*u'(x) => f'(x) = 4*(x^3 +...

Latest answer posted March 3, 2011 12:09 am UTC

Calculus

The required limit that x tends to is not specified. I take it to be 0 as substituting x = 0, makes the expression indefinite with 0/0. We can use the l'Hopital's rule here and substitute the...

Latest answer posted March 11, 2011 10:38 pm UTC

Calculus

We have to find the value of lim x--> 60 degrees[ 4cos x - 6(sin x)^2 + 3*tan x)^2] Here we see that cos 60 , (sin 60)^2 and (tan 60)^2 are defined, so we can just substitute x with 60....

Latest answer posted February 18, 2011 8:50 pm UTC

Calculus

We have to find the definite integral of y= - cos 3x * sin x, in the limit x = 0 to x = pi. Int [ - cos 3x * sin x] => Int [ cos 3x * sin dx] => Int [ (4(cos x)^3 - 3 cos x) sin x dx] let cos...

Latest answer posted February 15, 2011 2:09 am UTC

Calculus

F(s) = s + 2/s + 10. To find the maximum value of F(s) in the interval [1,4], we differentiate F(s) and equate the derivative to 0. Also, at the point of maximum value F''(s) is negative. F(s) =...

Latest answer posted March 3, 2011 1:26 am UTC

Calculus

We have to find the derivative of y = (10 + lg (x^10) + e^10x)^10. We use the chain rule to find the derivative of y. y' = 10 * (10 + lg (x^10) + e^10x)^9 * (10 / x + 10*e^10x) => 10 * (10 + lg...

Latest answer posted February 20, 2011 12:34 am UTC

Calculus

We have to find the value of lim x-->2 [ ln (x - 1)/(x - 2)] substituting x = 2, we get the indeterminate form 0/0; that permits the use of l'Hopital's Theorem and we can substitute the...

Latest answer posted March 3, 2011 1:50 am UTC

Calculus

The function f(x) is defined such that f(x) = x + 2 , if x <= 1 and f(x) = x^2, if x > 1. At the point x = 1, if we approach from the left lim x--> 1- [ f(x)] = 1 + 2 = 3. If we approach...

Latest answer posted March 12, 2011 12:21 am UTC

Calculus

We have to find the derivative of y = (x+3)^2 + 2(x+3)(x-4) + (x-4)^2 y = (x+3)^2 + 2(x+3)(x-4) + (x-4)^2 Use the relation [x^n]' = n*x^(n - 1) and the chain rule y' = 2( x + 3) + 2*( x + 3) + 2* (...

Latest answer posted February 13, 2011 12:42 am UTC

Calculus

We have to find the value of lim x--> inf+ [sin x / sqrt (x^2 + 1)] sin x is a periodic function the value of which oscillates between -1 and 1 as x tends to infinity. It is not possible to...

Latest answer posted February 18, 2011 9:45 pm UTC

Calculus

The area bound by the curve y = cosx/(sin^2x-4), the x axis and the lines x = 0 and x = pi/2 is the integral of y between the limits x = 0 and x = pi/2. y = cos x/ ((sin x)^2 - 4) let sin x = y...

Latest answer posted April 16, 2011 2:52 am UTC

Calculus

We have the functions f(x) = 3x+ 2 and g(x) = x^2 + 1 u = fog ( x) = f(g(x)) => f(x^2 + 1) => 3(x^2 + 1) + 2 => 3x^2 + 3 + 2 => 3x^2 + 5 v = gof(x) = g(f(x)) => g( 3x + 2) => (3x...

Latest answer posted February 20, 2011 12:52 am UTC

Calculus

We have the fraction (4x^2-1)/(4x^2-4x+1) and we have to find its derivative. (4x^2-1)/(4x^2-4x+1) Use the relation x^2 - y^2 = (x - y)(x + y) and (x -y)^2 = x^2 + y^2 - 2xy => (2x - 1)(2x + 1)...

Latest answer posted February 13, 2011 12:53 am UTC

Calculus

To find the minimum value of y = 4x^2 - 8x -18, we need to find the first derivative and equate it to 0. We solve for x and determine the value of the function for the value of x that we find. y =...

Latest answer posted February 20, 2011 1:35 am UTC

Calculus

We need to find f(x) given that f'(x) = sin 2x /((sin x)^2 - 4) let ((sin x)^2 - 4) = y dy = 2*sin x * cos x dx =>dy = sin 2x dx Int [ sin 2x /((sin x)^2 - 4) dx] => Int [ 1/y dy] => ln...

Latest answer posted April 16, 2011 3:13 am UTC

Calculus

We have to find the derivative of f(x)=(2x^2+1)/(2x^2-1) We use the quotient rule here, which gives the derivative of f(x)/g(x) as (f'(x)*g(x)* - f(x)*g'(x))/(g(x))^2 f(x)=(2x^2+1)/(2x^2-1) f'(x) =...

Latest answer posted February 17, 2011 5:16 pm UTC

Calculus

We have y = (3x^4+4x^3+1)(3x^2+2x^3-1) To find the derivative dy/dx we use the product rule. y' = (3x^4+4x^3+1)'(3x^2+2x^3-1) + (3x^4+4x^3+1)(3x^2+2x^3-1)' y' = (12x^3+12x^2)'(3x^2+2x^3-1) +...

Latest answer posted February 17, 2011 5:27 pm UTC

Calculus

We have to find the integral of y=4x^t + 5x^-1 + 6sin x. I assume t to be a constant. Int[ y dx] => Int[ 4x^t + 5x^-1 + 6sin x dx] => Int[4x^t dx] + Int[5x^-1 dx]+ Int [6sin x dx] =>...

Latest answer posted May 3, 2011 12:40 pm UTC

Calculus

We have to find the anti derivative of f(x) = (ln(2x-5)) / (2x-5). This can be done using substitution Int [ (ln(2x-5)) / (2x-5) dx] let ln(2x - 5) = t => dt/dx = 2/(2x - 5) => dt / 2 =...

Latest answer posted May 7, 2011 3:44 pm UTC

Calculus

We have to verify the value of lim x--> 1[ (x^2 - 6x + 5)/(x - 1)] Substituting x = 1 we get the indeterminate form 0/0. So, we can use l'Hopital's rule and substitute the numerator and...

Latest answer posted February 21, 2011 1:54 am UTC

Calculus

First I(a) needs to be determined. That is given as the definite integral of (2x - 3)*e^x for x = -a to x= 0 Int [ (2x - 3)*e^x dx] use integration by parts. Int[ u dv ] = u*v - Int [ v du] 2x - 3...

Latest answer posted May 7, 2011 4:32 pm UTC

Calculus

We have to find the derivative of y=(3+e^x+8x^3)^2*(2x+e^2x) We can use the product rule here y=(3+e^x+8x^3)^2*(2x+e^2x) y' = [(3+e^x+8x^3)^2*(2x+e^2x)]' =>y' = [(3+e^x+8x^3)^2]*(2x+e^2x)]' +...

Latest answer posted February 16, 2011 1:57 am UTC

Calculus

The function f(x) = e^(ln((x)^x)) taking the log to the base e of both the sides we get: ln f(x) = ln x^x differentiate both the sides f'(x)/f(x) = [ln x^x]' => f'(x)/f(x) = [x*ln x]' =>...

Latest answer posted April 1, 2011 2:22 am UTC

Calculus

As you have not specified what x is tending to, I take it as 2. We have to find the value of lim x--> 2 [(x^2-3x+2)/(x^2-4)] If we substitute x = 2, we get the indeterminate form 0/0. So we use...

Latest answer posted February 23, 2011 1:14 am UTC

Calculus

The function given is f(x) = 0.25x^4 + 3x^3 – 18x^2+ 10 At the extreme points f'(x) = 0 f'(x) = 4*0.25x^3 + 9x^2 - 36x x^3 + 9x^x - 36 = 0 => x(x^2 + 9x - 36) = 0 => x(x^2 + 12x - 3x - 36) =...

Latest answer posted April 18, 2011 2:28 am UTC

Calculus

We have the function f(x) =2x^3 + e^2x + sin 2x - ln x. We have to find f''(x). f'(x) = 6x^2 + 2e^2x + 2 cos 2x - 1/x f''(x) = 12x + 4e^2x - 4 sin 2x + 1/x^2 The required second derivative f(x) =...

Latest answer posted February 23, 2011 2:27 am UTC

Calculus

We have the function h(x) = 5x + (4/x). The critical points of this function are at points where h'(x) = 0 h'(x) = 5 - 4/x^2 5 - 4/x^2 = 0 => x^2 = 4/5 => x= 2/sqrt 5 and -2/sqrt 5 h''(x) =...

Latest answer posted March 21, 2011 4:26 pm UTC

Calculus

We have to find the anti derivative of y = sin x*(cos x)^n let cos x = y => dy/dx = -sin x => (-1)*dy = sin x dx Int [ sin x*(cos x)^n dx] => Int [ (-1)* y^n] => (-1)* y^(n+1) / (n +...

Latest answer posted March 20, 2011 2:42 am UTC

Calculus

We have the expression: tan( ln (sqrt (4e^x+2x^2)))^2 + cot (ln (sqrt (4e^x + 2x^2)))^2 We have to find the derivative of the expression. We use the chain rule and start from the innermost...

Latest answer posted February 3, 2011 11:54 pm UTC

Calculus

We have to find the derivative of y = ln[(x+1)^4*(x+7)^2*(x+2)^3] First simplify the expression using the properties of logarithms: log(a*b) = log a + log b and log(a^b) = b*log a y =...

Latest answer posted April 6, 2011 2:51 pm UTC

Calculus

As you have not specified what x tends to, I am taking it to be 0. The value of lim x-->0 [(3x+2)/(2x+1)]^[(2x+1)/(x+4)] is required. lim x-->0 [(3x+2)/(2x+1)]^[(2x+1)/(x+4)] substitute x =...

Latest answer posted April 29, 2011 11:47 pm UTC

Calculus

We have to prove that [(x+1)*lnx]/2 > x-1 [(x+1)*lnx]/2 > x-1 => [(x+1)*lnx] > 2(x-1) => ln x > 2(x-1)/(x+1) => ln x - 2(x-1)/(x+1) > 0 If we find the derivative of ln x -...

Latest answer posted January 21, 2011 9:47 pm UTC

Calculus

We have to prove that the derivative of arc tan (1-x^2) + arc cot (1-x^2) is 0. We know that the derivative of arc tan x = 1/[1 + (1-x^2)^2] and the derivative of arc cot x = -1 / [1 +...

Latest answer posted February 4, 2011 12:01 am UTC

Calculus

The function given has two variables x and y. We have to find the partial derivative f'x, which implies that y can be considered as a constant. f(x,y) = x^2*2^(x*y) f'x = [x^2]'*2^(x*y) +...

Latest answer posted March 20, 2011 4:02 am UTC

Calculus

The first derivative of a curve at any point gives the slope of the tangent at that point. Here the curve is defined by f(x) = x^2 - 2x + 2 f'(x) = 2x - 2 As the tangent is perpendicular to the the...

Latest answer posted January 23, 2011 10:17 pm UTC

What is dy/dx if square root x = 1+x^2*y ? The method required to answer this question is called implicit differentiation: d/dx (x)^0.5 =d/dx (1+x^2*y) (notice I converted the square root of x...