-
Calculus
We have the function y = e^ln(x^x) take the log of both the sides ln y = ln (e^ln(x^x)) => ln y = ln(x^x) => ln y = x*ln x differentiate both the sides (1/y)dy/dx = ln x + x/x dy/dx = y*(ln x...
-
Calculus
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...
-
Calculus
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 +...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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-...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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
-
Calculus
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...
-
Calculus
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...
-
Calculus
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 =...
-
Calculus
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...
-
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...
-
Calculus
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...
-
Calculus
We have to find the value of (x^2+2x-3)/|x-1| as x approaches from the left. As x approaches from the left x - 1 is always negative, so we have |x - 1| = (1 - x) lim x--> 1 [ (x^2+2x-3)/(1 -...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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)]...
-
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...
-
Calculus
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 -...
-
Calculus
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 =...
-
Calculus
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 =...
-
Calculus
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 =>...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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...
-
Calculus
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)*...
-
Calculus
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...
-
Calculus
To find the slant asymptote of x^3 / (x + 2)^2 we have to divide x^3 by (x + 2)^2 (x^2 + 4x + 4) | x^3...........................................| x - 4 ...........................x^3 + 4x^2 + 4x...
-
Calculus
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...
-
Calculus
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...
-
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...
-
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 =>...
-
Calculus
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...
-
Calculus
We have to find the integral of 1/(16x^2+24x+9) 1/(16x^2+24x+9) => 1/(4x+3)^2 let 4x + 3 = u => du/dx = 4 => dx = du/4 Int[ 1/(16x^2+24x+9) dx] => Int [ 1/u^2 du/4] => (1/4) Int [...
-
Calculus
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...
-
Calculus
We have f(x) = (3x + 1)*e^-x We use the product rule to find f'(x) f'(x) = (3x + 1)'*e^-x + (3x + 1)*(e^-x)' => 3*e^-x - (3x +1)e^-x => 3*e^-x - f(x) f''(x) = -3e^-x - f'(x) => -3e^-x -...
-
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...
-
Calculus
We have to find lim x--> -2 [(x^2-2x-8)/(x^3+8)] using l'Hopital's rule. First we find out if the l'Hopital's Rule can be used here. Substituting x = -2 we get the indeterminate form 0/0;...
-
Calculus
We have the function: f(x)=12x^4+24x^2+56 f(x) = 12x^4 + 24x^2 + 56 f'(x) = 48x^3 + 48x If f'(x) = 0 => 48x^3 + 48x = 0 => x^3 + 3 = 0 => x( x^2 + 1) = 0 x1 = 0 x2 = -sqrt (-1) => x2 =...
-
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...
-
Calculus
We have to find the anti-derivative of [ cos 2x - (cos x)^2]^-1 f(x) = [ cos 2x - (cos x)^2]^-1 => f(x) = [(cos x)^2 - (sin x)^2 - (cos x)^2]^-1 => f(x) = -(sin x)^-2 => f(x) = -1 / (cosec...
-
Calculus
To find the maximum and minimum values of the function f(x) = 2x^3 - 4x + 3, we need to find the first derivative. f(x) = 2x^3 - 4x + 3 f’(x) = 6x^2 - 4 Equate this to zero and solve for x 6x^2 -...