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A: Use formula of equation of tangent line
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A: Given- Point 8,-8. To Find- The slope of the tangent line at the given point. Identity Used- tanθ=yx
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A: Answer and explanation is given below
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A: Solve, as follows: 4xcosy+1=541cosy+x-sinydydx=0dydx=cosyxsiny
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Q: Find equations of the tangent line and normal line to the given curve at the specified point.
A: y=xx+1y'=12x(x+1)-x(1)(x+1)2y'=12x(x+1)-x(x+1)2At x=4, y'=124(4+1)-4(4+1)2=54-225=-3425=-3100
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A: Use differentiation for finding the slope.
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Q: What is a tangent line?
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- Determine the slope of the tangent line to the curve y=2 – x2 at the point (3,-7)Please find the slope of the tangent line to the curve defined by 3x^4+2xy−y^3=63 at the point (3,6).How many tangent lines to the curve y = x/(x + 1) pass through the point (1, 2)? At which points do these tangent lines touch the curve?
- The slope of the tangent line to the curve y=2x^3 at the point (-4,-128) is ____. The equation of this tangent line can be written in y=mx+b where m=___ and b=___What is the slope of the tangent line to the graph of y = x^2 + 1 at the point (0, 1)?What is the slope of the tangent line to the curve x = 3siny at the point (0, 2π)?