Calculus

We have to find Int [1/ (1 + 4x^2) dx]. First substitute u = 2x => du /dx = 2 => du /2 = dx Now Int [1/ (1 + 4x^2) dx] => Int [(1/2)*(1/ (1+u^2) du] => (1/2)*Int [1/ (1 + u^2) du] Now...

The critical points are determined by differentiating the function and equating the derivative to 0. It is solved to determine x. f(x) = sin x + cos x f'(x) = cos x - sin x = 0 => cos x = sin...

We need to determine the integral of (cos x)^7 * sin x. Int [(cos x)^7 * sin x dx] let cos x = u => - du = sin x dx => Int [ -u^7 du] => -u^8 / 8 + C substitute u = cos x => - (cos...

You want the limit of y=(1-cos 2x)/x^2 while x approaches 0. y = (1-cos 2x)/x^2 => [1 - (1 - 2*(sin x)^2)]/x^2 => 2*(sin x)^2/x^2 => 2*(sin x / x)^2 lim x--> 0 (sin x / x) = 1 Using...

The area of the region bounded by the curve y = sqrt (x - 1), the y- axis and the lines y = 1 and y = 5 is the limited integral of the expression of x in terms of y, between y = 5 and y = 1. y =...

We have to differentiate f(x) = x*cos 2x f'(x) = x'*cos 2x + x*(cos 2x)' f'(x) = cos 2x + x*(-sin 2x)*2 f'(x) = cos 2x - 2x*(sin 2x) The required derivative of f(x) = x*cos 2x is f'(x) = cos 2x -...

We first determine the points where the curves y = 8 - x^2 and y = x^2, meet. 8 - x^2 = x^2 => x^2 = 4 => x = 2 , x = -2 Now we find the integral of 8 - x^2 - x^2 between the limits x = -2...

We have to prove that lim x-->0 [(a^x - 1)/x] = ln a First, if we substitute x = 0, we get the indeterminate form 0/0. This allows the use of l"Hopital's rule and we can substitute the numerator...

We have to find Int [e^2x * cos 3x dx] Here the best way to solve would be to use integration by parts. Int [u dv] = u*v – Int [v du] take u = e^2x, du = 2*e^2x dx dv = cos 3x dx, v = (1/3)* sin...

We have to find the value of lim h-->0[[(3+h)^2-9]/h] lim h-->0[[(3+h)^2-9]/h] => lim h-->0[[(3 + h - 3)(3 + h + 3)/(3 + h - 3)] cancel (3 + h - 3) => lim h-->0[[(3 + h + 3)]...

To find the curve we integrate the given dy/dx = 3x^2 - 2x. Int [ 3x^2 - 2x dx ] => 3*x^3 / 3 - 2x^2 / 2 + C => x^3 - x^2 + C As the curve passes through (2 , 5) 5 = 2^3 - 2^2 + C => 5 =...

We have to find the value of lim x--> 90[ (1- sin x)/(cos x)^2] substituting x = 90 degrees, we get the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and...

We need to find the value of lim x-->0 [ tan 4x / tan 2x] If we substitute x = 0, we get the indeterminate form 0/0. This allows us to use l'Hopital's rule and substitute the numerator and the...

The function f(x) = x^(sin x) Let y = f(x) = x^(sin x) Take the natural log of both the sides ln y = ln [ x^(sin x)] => ln y = sin x * ln x Differentiate both the sides with respect to x =>...

We have dy/dx = 4x^3 + 4x. dy/dx = 4x^3 + 4x => dy = (4x^3 + 4x) dx Integrate both the sides Int [ dy ] = Int [ (4x^3 + 4x) dx ] => y = 4x^4 / 4 + 4*x^2 / 2 => y = x^4 + 2*x^2 + C As the...

We have to determine the definite integral of y = sin 2x /sqrt(1 + (sin x)^4), x = 0 to x = pi/2 Int [ sin 2x /sqrt(1 + (sin x)^4) dx] let 1 + (sin x)^2 = y dy/dx = 2*sin x* cos x = sin 2x => dy...

We have to find the value of lim x--> 0[ (sin 5x - sin 3x)/x] if we substitute x = 0, we get the form 0/0, which allows us to use the l'Hopital's rule and substitute the numerator and the...

The company wishes to manufacture a box with a volume of 44 cubic feet that is open on top and is twice as long as it is wide. Let the length of the box that uses minimum amount of material be L....

We have to find the antiderivative of y = x / sqrt ( x^2 - 9) Let u = x^2 - 9 => du / dx = 2x => x dx = du/2 Int [ x / sqrt ( x^2 - 9) dx] => Int [(1/2)*(1/ sqrt u) du] => Int [ (1/2)*...

We have to determine the value of lim x--> 0[(cos x - cos 3x) / x*sin x If we substitute x = 0, we get (1- 1) / 0 = 0/0 As this is an indeterminate form we use l'Hopital's rule and replace the...

We have to determine lim x-->0 [(2x - sin 2x)/x^3] If we substitute x = 0, we get the indeterminate form 0/0, so we use the l'Hopital's Rule and substitute the numerator and denominator with...

We have the function y = sin x + cos 3x. The derivative of sin x is cos x and the derivative of cos x is -sin x. Also, for a function of the form y= f(g(x)), the derivative of y or y' is given by...

We are given that f(x)=1+2x^5/x^2 = 1 + 2x^3. We have to find: lim x -->1 [(f(x) - f(1))/(x-1)] => lim x -->1 [(1+ 2x^3 - 1 - 2)/(x-1)] => lim x -->1 [(2x^3 - 2)/(x-1)]\ => lim x...

First we need to determine the points of intersection between lnx and ln^2 x ==> ln x = ln^2 x ==> ln^2 x - ln x = 0 ==> lnx ( lnx -1) =0 ==> lnx = 0 ==> x = 1 ==> lnx-1 = 0...

We have to verify that lim x-->0 [ ln(1+x)/x] = 1. substituting x = 0, we get the indeterminate form 0/0, therefore we can use the l'Hopital's rule and substitute the numerator and denominator...

We have to find the value of the definite integral of x^2/sqrt (x^3 + 1) between the limits x = 2 and x = 3. First we determine the indefinite integral and then substitute the values x = 3 and x =...

We have to find the limit of f(x)=(sin x-cos x)/cos 2x for x--> 45 degrees. We know that cos 2x = (cos x)^2 - (sin x )^2 lim x--> 0 [(sin x-cos x)/cos 2x] => lim x--> 0 [(sin x-cos...

We have to integrate [1/ (y^2 + 8y + 20) dy] Int [1/( y^2 + 8y + 20) dy] => Int [ 1 / (y^2 + 8y + 16 + 4) dy] => Int [ 1/((y + 4)^2 + 2^2) dy] if u = y + 4 , dy = du => Int [ 1/ ( u^2 +...

We have to find the value of lim x--> pi[ sin 5x / sin x] We see that substituting x with pi gives us the form 0/0 which is indeterminate. We can use therefore use l'Hopital's rule and use the...

We have to find the value of lim x-->0+ [(x^x)^2)] lim x-->0+ [(x^x)^2)] => lim x-->0+ [e^(ln ((x^x)^2))] => lim x-->0+ [e^(ln ((x^2x))] => lim x-->0+ [e^(2x*ln x)] As the...

We need to find the value of lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/2*(cos x)^2 - 1) - sin 2x / cos x. Substituting x = pi/4, gives us an indeterminate value. lim x--> pi/4 [sin x/(1-...

We have y=(1+x^2)^3 We have to find dy/dx. We can use the chain rule here. dy/dx = 3(1 + x^2)^2*2x => dy/dx = 6x(1 + x^2)^2 The required result is 6x*(1 + x^2)^2

We have to find the area enclosed between y=x^2 - 2x + 2 and y = -x^2 + 6. First lets find the points of intersection x^2 - 2x + 2 = -x^2 + 6 => 2x^2 - 2x - 4 = 0 => x^2 - x - 2 = 0 => x^2...

We have the function f(x) = e^2x + (sin 2x) / 2x f'(x) = [e^2x + (sin 2x) / 2x]' => (e^2x)' + [(sin 2x/2x]' => 2*e^2x + (2x*2*cos 2x - 2*sin 2x )/4*x^2 => 2*e^2x + (2x *cos 2x - sin 2x...

The extreme values of a function occur at the points where the derivative is equal to 0. f(x) = 2x^3 + 3x^2 - 12x + 5 => f'(x) = 6x^2 + 6x - 12 6x^2 + 6x - 12 = 0 => x^2 + x - 2 = 0 => x^2...

We have to find the integral of y= x / sqrt(x^2+9) Int [ x/ sqrt (x^2 + 9) dx] let x^2 + 9 = u, du / 2 = x dx => (1/2)*Int [ 1/ sqrt u du] => (1/2)*Int [ u^(-1/2) du] => (1/2) u^(1/2) /...

We have to find the second derivative of y = e^5x + (ln x) / 2x y = e^5x + (ln x) / 2x => y = e^5x + (1/2)(ln x)*(1/x) y' = 5*e^5x + (1/2)[(-1)(x^-2)(ln x) + (1/x^2)] y'' = 25*e^5x +...

We have to find the derivative of y = arc sin x/(1-x^2). We use the quotient rule here: y' = [(arc sin x)'*(1 - x^2) - ( arc sin x)*(1 - x^2)']/(1 - x^2)^2 => [sqrt(1-x^2)*(1 - x^2) + 2x*(arc...

We have to find the extreme point of the curve y = 3x - 6x^2. To do that we find the first derivative of 3x - 6x^2 and equate it to zero. This is solved for x. Now y = 3x - 6x^2 y' = 3 - 12x 3 -...

We have to find the integral of e^ sqrt x / sqrt x. Let u = sqrt x => du/dx = 1/2* sqrt x => 2* du = dx / sqrt x Int [ (e^ sqrt x / sqrt x) dx] => Int [ e^u * (dx/sqrt x)] => Int [ e^u...

We have to find the value of lim x-->0 [(sin 2x - sin 6x) / 4x] We see that substituting x = 0 at this stage gives an indeterminate form 0/0. So we use l'Hopital's rule and substitute the...

We have the function g(x) = sin 9x. We have to find the second derivative of g(x). We use the chain rule here. g(x) = sin 9x => g'(x) = 9 cos 9x => g''(x) = -81 sin 9x The required second...

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) +...

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) =...

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 -...

We have to find the value of lim x--> 1[ (x^2+11x-12)/(x-1)]. As substituting x = 0 gives us the indeterminate form 0/0, we can use l'Hopital's theorem and substitute the numerator and the...

We have to find y' for y = (2-x)^(sqrt x) Use natural logariths for both the sides ln y = ln[ (2-x)^(sqrt x)] use the property ln a^x = a*ln x => ln y = (sqrt x)*ln ( 2 - x) Do implicit...

We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)] substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and...

Hi, djshan, Sorry, but I'm not too sure what you want us to do here. Are we going to graph this? Find the intercepts? Find the zeros? Something else? I would assume we are graphing it. To...

We have to find the derivative of (x - sqrt x) / (5x^2 - 3). As we have the expression as a quotient, we use the quotient rule. f(x) = u(x) / v(x) => f'(x) = (u'(x)*v(x) - u(x)*v'(x))/v(x)^2...