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Which of the following is equivalent to $\frac{\log_{c}(6)}{\log(6)}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\log(c)$ $\newline$ (B) $\log_{c}(1)$ $\newline$ (C) $\frac{1}{\log(c)}$ $\newline$ (D) $\frac{1}{\log(6)}$

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Find indefinite integrals using the power rule (Level 2)

Full solution

Q. Which of the following is equivalent to

\frac{\log_{c}(6)}{\log(6)}

?

\newline

Choose

1

answer:

\newline

(A)

\log(c)

\newline

(B)

\log_{c}(1)

\newline

(C)

\frac{1}{\log(c)}

\newline

(D)

\frac{1}{\log(6)}

Differentiate with respect to $x$ : Differentiate both sides of the equation with respect to $x$ using implicit differentiation. $\newline$ To differentiate $\tan(x+y)$ , we use the chain rule and the fact that the derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$ . For $\sec(x-y)$ , we also use the chain rule and the fact that the derivative of $\sec(u)$ with respect to $u$ is $\sec(u)\tan(u)$ . We must also remember to apply the derivative to the inside function ( $x$ $0$ or $x$ $1$ ) which will give us an additional factor of $x$ $2$ for each term since the derivative of $x$ with respect to $x$ is $x$ $2$ and the derivative of $x$ $6$ with respect to $x$ is $x$ $8$ ( $x$ $9$ since $x$ $6$ is a function of $x$ ). $\newline$ Differentiating $\tan(x+y)$ gives $\tan(x+y)$ $3$ because of the chain rule. Differentiating $\sec(x-y)$ gives $\tan(x+y)$ $5$ . The right side of the equation is a constant, so its derivative is $\tan(x+y)$ $6$ . $\newline$ So we have: $\newline$ $\tan(x+y)$ $7$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . We can rearrange the terms to isolate $\frac{dy}{dx}$ : $\sec^2(x+y) \cdot \frac{dy}{dx} - \sec(x-y)\tan(x-y) \cdot \frac{dy}{dx} = -\sec^2(x+y)$ Combining like terms gives us: $\frac{dy}{dx} \cdot (\sec^2(x+y) - \sec(x-y)\tan(x-y)) = -\sec^2(x+y)$ Now we can solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{-\sec^2(x+y)}{(\sec^2(x+y) - \sec(x-y)\tan(x-y))}$
Evaluate at $(\pi/8, \pi/8)$ : Evaluate $\frac{dy}{dx}$ at the given point $(\pi/8), (\pi/8)$ . We need to plug in $x = \pi/8$ and $y = \pi/8$ into our expression for $\frac{dy}{dx}$ : $\frac{dy}{dx} = -\sec^2((\pi/8) + (\pi/8)) / (\sec^2((\pi/8) + (\pi/8)) - \sec((\pi/8) - (\pi/8))\tan((\pi/8) - (\pi/8)))$ Since $(\pi/8) - (\pi/8) = 0$ , $\sec(0) = 1$ and $\tan(0) = 0$ , the expression simplifies to: $\frac{dy}{dx}$ $0$
Calculate slope of tangent line: Calculate the slope of the tangent line. $\newline$ We know that $\sec(\frac{\pi}{4}) = \sqrt{2}$ , so we can substitute this into our expression: $\newline$ $\frac{dy}{dx} = -\frac{2}{(2 - 1)}$ $\newline$ $\frac{dy}{dx} = -2$ $\newline$ This is the slope of the tangent line at the point $\left(\frac{\pi}{8}, \frac{\pi}{8}\right)$ .
Write tangent line equation: Write the equation of the tangent line. $\newline$ The equation of a line is $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is a point on the line. We have $m = -2$ and our point is $\left(\frac{\pi}{8}, \frac{\pi}{8}\right)$ , so the equation of the tangent line is: $\newline$ $y - \left(\frac{\pi}{8}\right) = -2\left(x - \left(\frac{\pi}{8}\right)\right)$
Simplify tangent line equation: Simplify the equation of the tangent line. $\newline$ We can distribute the $-2$ and move $\frac{\pi}{8}$ to the other side to get the final equation: $\newline$ $y = -2x + \frac{\pi}{4} + \frac{\pi}{8}$ $\newline$ $y = -2x + \frac{3\pi}{8}$ $\newline$ This is the equation of the tangent line at the given point.

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