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  • 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...

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  • 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...

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  • 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...

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  • 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...

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  • 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...

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  • 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 +...

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  • 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}}`...

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  • 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 =...

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  • 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`...

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  • 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...

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  • 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`...

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  • 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...

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  • 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...

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  • 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`...

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  • 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...

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  • 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`...

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  • 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...

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  • 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...

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  • 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 +...

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  • 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,`...

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  • 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 -...

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  • 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...

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  • 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...

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  • 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)...

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  • 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)...

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    1 educator answer.

  • 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' =...

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  • 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) -...

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  • 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) =...

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    1 educator answer.

  • 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))...

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  • 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)...

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  • 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)...

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    1 educator answer.

  • 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...

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  • 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)) =...

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    1 educator answer.

  • 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...

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  • 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...

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  • 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).`

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  • 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...

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  • 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...

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  • 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) =...

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  • 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:...

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  • 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...

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  • 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...

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  • 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 *...

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  • 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:...

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  • 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],`...

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  • 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...

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  • 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....

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  • 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...

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  • 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...

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  • Math
    `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)...

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