Which of the following is equivalent to (log_(c)(6))/(log(6)) ? Choose 1 answer: (A) log(c) (B) log_(c)(1) (C) (1)/(log(c)) (D) (1)/(log(6))

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Differentiate with respect to x: Differentiate both sides of the equation with respect to x using implicit differentiation. 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+y or x-y) which will give us an additional factor of 1 for each term since the derivative of x with respect to x is 1 and the derivative of y with respect to x is dy/dx (y' since y is a function of x). Differentiating tan(x+y) gives sec^2(x+y) * (1 + dy/dx) because of the chain rule. Differentiating sec(x-y) gives sec(x-y)tan(x-y) * (1 - dy/dx). The right side of the equation is a constant, so its derivative is 0. So we have: sec^2(x+y) * (1 + dy/dx) + sec(x-y)tan(x-y) * (1 - dy/dx) = 0Solve for dy/dx: Solve for dy/dx. We can rearrange the terms to isolate dy/dx: sec^2(x+y) * dy/dx - sec(x-y)tan(x-y) * dy/dx = -sec^2(x+y) Combining like terms gives us: dy/dx * (sec^2(x+y) - sec(x-y)tan(x-y)) = -sec^2(x+y) Now we can solve for dy/dx: dy/dx = -sec^2(x+y) / (sec^2(x+y) - sec(x-y)tan(x-y))Evaluate at (pi/8, pi/8): Evaluate 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 dy/dx: 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: dy/dx = -sec^2((pi/4)) / (sec^2((pi/4)) - 1)Calculate slope of tangent line: Calculate the slope of the tangent line. We know that sec(pi/4) = sqrt(2), so we can substitute this into our expression: dy/dx = -2 / (2 - 1) dy/dx = -2 This is the slope of the tangent line at the point ((pi/8), (pi/8)).Write tangent line equation: Write the equation of the tangent line. The equation of a line is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. We have m = -2 and our point is ((pi/8), (pi/8)), so the equation of the tangent line is: y - (pi/8) = -2(x - (pi/8))Simplify tangent line equation: Simplify the equation of the tangent line. We can distribute the -2 and move (pi/8) to the other side to get the final equation: y = -2x + pi/4 + pi/8 y = -2x + 3pi/8 This is the equation of the tangent line at the given point.

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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