# Homework Help

eNotes Homework Help is a way for educators to help students understand their school work. Our experts are here to answer your toughest academic questions! Once it’s posted to our site, your question could help thousands of other students.

Filter Questions
• Math
For the given integral: int 3/(2sqrt(x)(1+x)) dx , we may apply the basic integration property: int c*f(x) dx = c int f(x) dx . int 3/(2sqrt(x)(1+x)) dx = 3/2int 1/(sqrt(x)(1+x)) dx . For...

Asked by enotes on via web

• Math
Recall that int f(x) dx = F(x) +C  where: f(x) as the integrand function F(x) as the antiderivative of f(x) C as the constant of integration.. For the given problem, the integral: int...

Asked by enotes on via web

• Math
We have to evaluate the integral: \int\frac{sinx}{7+cos^2x}dx Let cos x=u So, -sinxdx=du Hence we have, \int \frac{sinx}{7+cos^2x}dx=\int\frac{-du}{u^2+7} =-\int...

Asked by enotes on via web

• Math
We have to evaluate the integral : \int \frac{sec^2x}{\sqrt{25-tan^2x}}dx Let tanx =t So, sec^2x dx=dt Therefore we have, \int \frac{sec^2x}{\sqrt{25-tan^2x}}dx=\int...

Asked by enotes on via web

• Math
Given the integral: \int \frac{2}{x\sqrt{9x^2-25}}dx Let x=\frac{5}{3}sect  So, dx=\frac{5}{3}sect tant dt Hence we have, \int \frac{2}{x\sqrt{9x^2-25}}dx=\int \frac{\frac{10}{3}sect...

Asked by enotes on via web

• Math
The substitution u = 1/2 e^(2x) make this integral a well-known one: du = e^(2x) dx, so e^(2x) dx from the numerator is equal to du, and 4 + e^(4x) from the denominator is equal to 4 +...

Asked by enotes on via web

• Math
We have to evaluate the integral: \int \frac{1}{x\sqrt{1-(ln x)^2}}dx Let ln x=u So, \frac{dx}{x}=du Therefore we have, \int \frac{1}{x\sqrt{1-(ln(x))^2}}dx=\int \frac{du}{\sqrt{1-u^2}}...

Asked by enotes on via web

• Math
Make the substitution u = t^2/5, then du = (2t dt)/5, t dt = 5/2 du and t^4 = 25u^2. The integral becomes int (5/2)/(25u^2 + 25) du = 1/10 int (du)/(1 + u^2) = 1/10 arctan(u) + C =...

Asked by enotes on via web

• Math
For the given integral: int 1/(xsqrt(x^4-4))dx , we may apply u-substitution by letting: u =x^4-4  then  du = 4x^3 dx . Rearrange du = 4x^3 dx into (du)/( 4x^3)= dx Plug-in u =x^4-4...

Asked by enotes on via web

• Math
Indefinite integral are written in the form of int f(x) dx = F(x) +C where: f(x) as the integrand F(x) as the anti-derivative function C as the arbitrary constant known...

Asked by enotes on via web

• Math
Given the integral : \int \frac{1}{4+(x-3)^2}dx let x-3=t So, dx=dt therefore we have, \int \frac{1}{4+(x-3)^2}dx=\int \frac{1}{4+t^2}dt =\int \frac{1}{2^2+t^2}dt...

Asked by enotes on via web

• Math
Indefinite integral are written in the form of int f(x) dx = F(x) +C where: f(x) as the integrand F(x) as the anti-derivative function C as the arbitrary constant known...

Asked by enotes on via web

• Math
Recall that the indefinite integral is denoted as: int f(x) dx =F(x)+C There properties and basic formulas of integration we can apply to simply certain function. For the problem int...

Asked by enotes on via web

• Math
We have to evaluate the integral : \int \frac{dx}{x\sqrt{4x^2-1}} Let \sqrt{4x^2-1}=u So, \frac{1}{2\sqrt{4x^2-1}}.8x dx=du \frac{4xdx}{\sqrt{4x^2-1}}=du...

Asked by enotes on via web

• Math
We have to evaluate \int \frac{dx}{\sqrt{1-4x^2}} Let x=\frac{1}{2} sint  So, dx= \frac{1}{2}cost dt Hence we have, \int \frac{dx}{\sqrt{1-4x^2}}=\int \frac{\frac{1}{2}cost...

Asked by enotes on via web

• Math
We have to evaluate the integral: \int \frac{dx}{\sqrt{9-x^2}} let x=3sint So, dx=3cost dt Hence we have, \int \frac{dx}{\sqrt{9-x^2}}=\int \frac{3cost}{\sqrt{9-9sin^2t}}dt...

Asked by enotes on via web

• Math
First, check that the given point satisfies the equation: arctan(1 + 0) = 0 + pi/4 is true. The slope of the tangent line is y'(x) at the given point. Differentiate the equation with respect to...

Asked by enotes on via web

• Math
Does the given point belongs to the given curve? x^2 + x arctany = pi^2/16 - pi/4*pi/4 = 0, and y - 1 = 0 also. Then equation of the tangent line is (y - 1) = (x + pi/4)*y'(-pi/4). Perform...

Asked by enotes on via web

• Math
This function is defined on entire real axis and is differentiable everywhere. Its derivative is f'(x) = 1/(1+x^2) - 1/(1 + (x-4)^2) = (1 + (x-4)^2 - 1 - x^2)/((1+x^2)(1 + (x-4)^2)) = = (-8x +...

Asked by enotes on via web

• Math
This function is defined on [-1, 1] and is differentiable on (-1, 1). Its derivative is f'(x) = 1/sqrt(1-x^2) - 2. The derivative doesn't exist at x = +-1. It is zero where 1-x^2 = 1/4,...

Asked by enotes on via web

• Math
Check that the given point belongs to the given curve: pi/4 = 3*1/2*arcsin(1/2) is true, because arcsin(1/2) = pi/6 and 3/2*pi/6 = pi/4. The tangent line has an equation (y - pi/4) = (x -...

Asked by enotes on via web

• Math
The equation of the tangent line has the form (y-y_0) = (x-x_0)*y'(x_0). Here x_0 = 1 and y_0 = 2 pi. This point is really on the graph, because y_0 = 2pi is really equal to 4arccos(x_0...

Asked by enotes on via web

• Math
Equation of a tangent line to the graph of function f at point (x_0,y_0) is given by y=y_0+f'(x_0)(x-x_0). The first step to finding equation of tangent line is to calculate the derivative...

Asked by enotes on via web

• Math
The derivative of y with respect to is denoted as y' or (dy)/(dx) . For the given equation: y = arctan(x/2) -1/(2(x^2+4)) , we may apply the basic property of derivative: d/(dx) (u-v)...

Asked by enotes on via web

• Math
Recall that the derivative of y with respect to is denoted as y' or (dy)/(dx) . For the given equation: y = arctan(x) +x/(1+x^2) , we may apply the basic property of derivative: d/(dx)...

Asked by enotes on via web

• Math
y=25arcsin(x/5) - xsqrt(25-x^2) Before taking the derivative, express the radical in exponent form. y=25arcsin(x/5) - x(25-x^2)^(1/2) To get y', take the derivative of each term. y' =...

Asked by enotes on via web

• Math
This derivative requires the product rule, the chain rule, several table derivatives etc: y'(x) = 8*(1/4)/sqrt(1-x^2/16) - sqrt(16-x^2)/2 - x*(-2x)/(4sqrt(16-x^2)) = = 8/sqrt(16-x^2) -...

Asked by enotes on via web

• Math
The derivative of y in terms of x is denoted by (dy)/(dx) or y’' For the given problem: y = xarctan(2x) -1/4ln(1+4x^2) , we may apply the basic differentiation property: d/(dx) (u-v) =...

Asked by enotes on via web

• Math
Differentiating this function includes use of linearity, the product rule and the chain rule: y'(x) = (x*arcsin(x))' + (sqrt(1-x^2))' =  =arcsin(x) + x/sqrt(1-x^2) + (-2x)/(2sqrt(1-x^2))...

Asked by enotes on via web

• Math
The derivative of y in terms of x is denoted by d/(dx)y or y' . For the given problem: y =1/2[xsqrt(4-x^2)+4arcsin(x/2)] , we apply the basic derivative property: d/(dx) c*f(x) = c d/(dx)...

Asked by enotes on via web

• Math
The derivative of y in terms of x is denoted by (dy)/(dx) or y’ . For the given problem: y = 1/2(1/2ln((x+1)/(x-1)) +arctan(x)) , we may apply the basic differentiation property: d/(dx)...

Asked by enotes on via web

• Math
We need the derivatives of ln(u) and arctan(u), they are 1/u and 1/(1 + u^2), and the chain rule. The result is y'(t) = (2t)/(t^2+4) - 1/2*(1/2)/(1+t^2/4) =(2t)/(t^2+4) - 1/(t^2 + 4) = (2t...

Asked by enotes on via web

• Math
y = 2arccos(x) - 2sqrt(1-x^2) First, express the radical in exponent form. y = 2arccos(x)-2(1-x^2)^(1/2) To take the derivative of this, use the following formulas: d/dx(arccos(u)) =...

Asked by enotes on via web

• Math
Recall that the derivative of a function f at a point x is denoted as f'(x). The given function: f(x)= arcsin(x)+arccos(x) has inverse trigonometric terms. There are basic formulas for the...

Asked by enotes on via web

• Math
This function is a composite one, it may be expressed as h(t) = u(v(t)), where v(t) = arccos(t) and u(y) = sin(y). The chain rule is applicable here, h'(t) = u'(v(t))*v'(t). This gives us...

Asked by enotes on via web

• Math
We use the product rule, (uv)' = u'v + uv', for u = x^2 and v = arctan(5x), and then the chain rule: h'(x) = 2x*arctan(5x) + x^2 (arctan(5x))' = = 2x*arctan(5x) + (5 x^2)/(1 + 25x^2).

Asked by enotes on via web

• Math
The given function g(x) has a form u(x)/v(x), thus its derivative may be determined by the quotient rule: (u/v)' = (u' v - u v')/v^2. Also we need to know that the derivative of acrsine...

Asked by enotes on via web

• Math
The given function f(x) = arctan(sqrt(x)) is in a inverse trigonometric form. The basic derivative formula for inverse tangent is: d/(dx) arctan(u) = ((du)/(dx))/sqrt(1-u^2) . Using...

Asked by enotes on via web

• Math
The given function: f(x) =arctan(e^x) is in a form of inverse trigonometric function. It can be evaluated using the derivative formula for inverse of tangent function: d/(dx)arctan(u) =...

Asked by enotes on via web

• Math
The given function: f(x) =arcsec(2x)  is in a form of an inverse trigonometric function. For the derivative formula of an inverse secant function, we follow:...

Asked by enotes on via web

• Math
To take the derivative of the given function: g(x) =3arccos(x/2) , we can apply the basic property: d/(dx) [c*f(x)] = c * d/(dx) [f(x)] . then g'(x) = 3 d/(dx) (arccos(x/2)) To solve for...

Asked by enotes on via web

• Math
The derivative of a function with respect to t is denoted as f'(t) The given function: f(x) = arcsin(t^2)  is in a form of a inverse trigonometric function. Using table of derivatives, we have...

Asked by enotes on via web

• Math
The derivative of function f with respect to x is denoted as f'(x) . To take the derivative of the given function: f(x) =2arcsin(x-1) , we can apply the basic property: d/(dx) [c*f(x)] = c *...

Asked by enotes on via web

• Math
arcsin(-x)=-arcsin(x) Let y=arcsin(-x) sin(y)=sin(arcsin(-x)) sin(y)=-x -sin(y)=x sin(-y)=x arcsin(sin(-y))=arcsin(x) -y=arcsin(x) y=-arcsin(x) Therefore:...

Asked by enotes on via web

• Math
The function at the left side, arccos(x), is defined at [-1, 1], because it is the range of cos(x). The function at the right side, arcsec(x), is defined at (-oo, -1] uu [1, +oo],...

Asked by enotes on via web

• Math
arcsin (sqrt(2x)) = arccos(sqrtx) To solve, let's consider the right side of the equation first. Let it be equal to theta. theta = arccos(sqrtx) Then, express it in terms of cosine. cos...

Asked by enotes on via web

• Math
The range of arctan function is (-pi/2, pi/2), on which interval tangent function is strictly monotone. Therefore we may take tangent on both sides, and the solution set will remain the same....

Asked by enotes on via web

• Math
The range of arcsine function is [-pi/2, pi/2], 1/2 belongs to this interval and sine is monotone on this interval. Therefore we may apply sine for both parts and don't lose any solutions or get...

Asked by enotes on via web

• Math
Given: sec(arctan(-3/5)) tany=-3/5 when y is in the domain -pi/2<y<0 Imagine a right triangle with opposite side having a length of 3 and adjacent side having a length of 5. The...

Asked by enotes on via web

cot(arcsin (-1/2)) To evaluate this, let's first consider the innermost expression. Let it be equal to y. y = arcsin(-1/2) Rewriting it in terms of sine function, the equation becomes `sin (y)...