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  • Math
    Given `\frac{dy}{dx}=\frac{1}{\sqrt{4-x^2}}, y(0)=\pi` , we have to find y. So we can write, `dy=\frac{dx}{\sqrt{4-x^2}}` Integrating both sides we have, `y=\int \frac{dx}{\sqrt{4-x^2}}+C` Now...

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  • Math
    Complete the square at the denominator: `x^4 + 2x^2 + 2 = (x^2)^2 + 2x^2 + 1 + 1 = (x^2 + 1)^2 + 1.` Now we see the substitution `y = x^2 + 1,` then `dy = 2x dx,` and the integral becomes `int...

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  • Math
    By completing the square and making simple substitution, we will reduce this integral to a table one. `-x^2-4x = -(x^2 + 4x + 4) + 4 = -(x+2)^2 + 4 = 4 - (x+2)^2.` Now make a substitution `y =...

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  • Math
    Recall that `(arccos(x))' = -1/sqrt(1-x^2)` and make the substitution `y = arccos(x),` then `dy = -1/sqrt(1-x^2).` The limits of integration are from `arccos(0) = pi/2` to `arccos(1/sqrt(2)) =...

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  • Math
    We have to evaluate the definite integral: `\int_{0}^{1/\sqrt{2}}\frac{arc sinx}{\sqrt{1-x^2}}dx` Let arc sinx=t Differentiating both sides we get, `\frac{1}{\sqrt{1-x^2}}dx=dt`...

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  • Math
    Make the substitution `u = sin(x),` then `du = cos(x) dx.` The integration limits for `u` are from `sin(0) = 0` to `sin(pi/2) = 1,` and the integral becomes `int_0^1 (du)/(1 + u^2) = arctan(1) -...

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  • Math
    Make a substitution `y = 2x,` then `dy = 2 dx` and the limits of integration for `y` are from `0` to `3.` The integral becomes the table one: `int_0^3 1/(1+y^2) (dy)/2 = 1/2 (tan)^(-1)(y) |_(y=0)^3...

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  • Math
    We have to evaluate the integral : `\int_{0}^{\sqrt{2}}\frac{dx}{\sqrt{4-x^2}}` let `x=2sin t` So, `dx=2cos t dt` When x=0, t=0 `x=\sqrt{2}, t=\pi/4` So we have,...

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  • Math
    Make a substitution `x = 3y,` then `dx = 3 dy` and the limits of integration for y are from 0 to 1/18. The integral becomes a table one: `int_0^(1/18) (9 dy)/(sqrt(9-9y^2)) = int_0^(1/18) (3...

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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
    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 `\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
    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. We can solve for the derivative of...

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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 `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 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
    `arccos(1)` Let this expression be equal to y. `y = arccos(1)` Rewriting it in terms of cosine function, the equation becomes: `cos(y) = 1` Base on the Unit Circle Chart, cosine is 1 at angles 0...

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  • Math
    `arccos(1/2)` Let this expression be equal to y. `y = arccos(1/2)` Rewriting this in terms of cosine function the equation becomes: `cos(y) =1/2` Base on the Unit Circle Chart, cosine is 1/2 at...

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  • Math
    `arcsin(0)` Let this expression be equal to y. `y =arcsin(0)` Re-writing this equation in terms of sine function, it becomes: `sin (y) = 0` Base on the Unit Circle Chart (see attached figure), sine...

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  • Math
    `arcsin(1/2)` Let this expression be equal to y. `y =arcsin(1/2)` Rewriting it in terms of sine function, the equation becomes: `sin(y) = 1/2` Base on the Unit Circle Chart (see attached figure),...

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  • Math
    To evaluate the integral: `int_(-4)^(4) 3^(x/4) dx` , we follow the formula based from the First Fundamental Theorem of Calculus: `int_a^bf(x)dx=F(b)- F(a)` wherein f is a continuous and F is...

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  • Math
    Recall the First Fundamental Theorem of Calculus: If f is continuous on closed interval [a,b], we follow: `int_a^bf(x)dx` = F(b) - F(a) where F is the anti-derivative of f on [a,b]. This...

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  • Math
    By definition, if the function F(x) is the antiderivative of f(x) then we follow the indefinite integral as `int f(x) dx = F(x)+C` where: f(x) as the integrand F(x) as the...

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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
    The point is really on the graph, because `2^(-(-1)) = 2.` The tangent line has the slope of `y'(-1),` and the equation `y - 2 = y'(-1)(x+1).` It is clear that `y'(x) = -2^(-x) ln2` and `y'(-1) =...

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  • Math
    Derivative of a function h with respect to t is denoted as h'(t). The given function: `h(t) = log_5(4-t)^2` is in a form of a logarithmic function. From the derivative for logarithmic functions,...

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  • Math
    `y= log_3(x^2-3x)` The derivative formula of a logarithm is: `d/(dx) [log_a (u)] = 1/(ln(a) * u) * (du)/(dx)` Applying this formula, the derivative of the function will be: `(dy)/(dx) = d/(dx)...

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  • Math
    `y=log_4(5x + 1)` The derivative formula of a logarithm is: `d/(dx) [log_a (u)] = 1/(ln(a) * u) * (du)/(dx)` Applying that formula, the derivative of the function will be: `(dy)/(dx) = d/(dx)[...

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  • Math
    `f(t) = 3^(2t)/t` To take the derivative of this function, use the quotient rule `(u/v)'= (v*u' - u*v')/v^2`. Applying that, f'(t) will be: `f'(t) = (t * (3^(2t))' - 3^(2t)*(t)')/t^2` `f'(t) =...

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  • Math
    The derivative of a function f at a point x is denoted as y' = f'(x). There are basic properties and formula we can apply to simplify a function such as the Product Rule provides the formula:...

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  • Math
    Recall that the derivative of a function f at a point x is denoted as `y' = f'(x)` . There basic properties and formula we can apply to simplify a function. For the problem `y = x(6^(-2x)),` we...

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  • Math
    The given function: `f(x)=x9^x ` has two factors since that is the same as `f(x) =x* 9^x` . In this form, we can apply the Product Rule for derivative. Product Rule provides the formula: `f(x) =...

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  • Math
    `y=6^(3x-4)` The derivative formula of an exponential function is: `d/(dx) (a^u) = ln(a) * a^u * (du)/dx` Applying this formula, the derivative of a function will be: `(dy)/(dx) = d/(dx)...

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  • Math
    `y=5^(-4x)` The derivative formula of an exponential function is: `d/(dx) (a^u) = ln(a) * a^u * (du)/(dx)` Applying this formula, the derivative of the function is: `(dy)/(dx) = d/(dx)(5^(-4x))`...

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  • Math
    The derivative of f(x) with respect to x is denoted a f'(x). The given function f(x) = 3^(4x) is in exponential form which means we can apply the basic integration formula: `d/(dx)(a^u) =...

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  • Math
    By definition, the derivative of f(x) with respect to x is denoted a f'(x) where `f'(x) = lim (f(x+h) -f(x))/h ` as` h->0` . Instead of using the limit of difference quotient, we may apply the...

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  • Math
    To simplify the logarithmic equation: `log_5(sqrt(x-4))=3.2` , recall the logarithm property: `a^((log_(a)(x))) = x` . When a logarithm function is raised by the same base, the log cancels out...

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  • Math
    To solve a logarithmic equation, we may simplify or rewrite it using the properties of logarithm. For the given problem `log_(3)(x^2)=4.5` , we may apply the property: `a^((log_(a)(x))) = x` The...

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  • Math
    In solving a logarithmic equation, we may simplify using logarithm properties. Recall the logarithm property: `a^((log_(a)(x))) = x` . When we raise the log with the same base, the "log" will...

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  • Math
    `log_2 (x-1) =5` To solve, convert the equation to exponential form. Take note that if a logarithmic equation is in the form `y = log_b (x)` its equivalent exponential equation is `x= b^y` So...

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  • Math
    Problem:`3(5^(x-1))=86` ` ` To simplify, we divide both sides by 3: `(3(5^(x-1)))/3=(86)/3` `5^(x-1)=(86)/3` ` ` ` ` Take the "log" on both sides to apply the logarithm property:...

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  • Math
    For exponential equation:`2^(3-z)=625` , we may apply the logarithm property: `log(x^y) = y * log (x)` . This helps to bring down the exponent value. Taking "log" on both sides:...

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  • Math
    For exponential equation: `5^(6x)= 8320` , we may apply the logarithm property: `log(x^y) = y * log (x).` This helps to bring down the exponent value. Taking "log" on both sides:...

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  • Math
    Problem:` 3^(2x)=75` is an exponential equation. To simplify, we need to apply logarithm property: `log(x^y) = y*log(x)` to bring down the exponent that is in terms of x. Taking "log" on both...

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  • Math
    `log_3(x) + log_3(x - 2) = 1` The logarithms at the left side have the same base. So express the left side with one logarithm only using the rule `log_b (M) + log_b (N) = log_b(M*N` ). `log_3(x *...

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  • Math
    `x^2-x=log_5 (25)` First, simplify the right side of the equation. To do so, factor 25. `x^2 - x = log_5 (5^2)` Then, apply the logarithm rule `log_b (a^m) = m * log_b (a)` . `x^2 - x = 2 * log_5...

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  • Math
    `log_b (27) = 3` To solve, convert this to exponential equation. Take note that if a logarithmic equation is in the form `y = log_b (x)` its equivalent exponential equation is: `x = b^y` So...

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  • Math
    `log_3 (x) = -1` To solve, convert this to exponential equation. Take note that if a logarithmic equation is in the form `y = log_b (x)` its equivalent exponential equation is: `x = b^y` So...

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