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The slope of the tangent to the graph ln(x+y)=x^2-y-89 at the point (9,-8) is the value of the first derivative dy/dx at that point.
[ln(x+y)]' = [x^2 - y - 89]'
=> (1 + dy/dx)/(x + y) = 2x - dy/dx
=> 1 + dy/dx = (2x - dy/dx)(x + y)
=> 1 + dy/dx = 2x(x + y) - dy/dx(x + y)
=> dy/dx[ 1 + x + y] = 2x(x + y) - 1
=> dy/dx = [2x(x + y) - 1]/[ 1 + x + y]
At the point (9, -8)
dy/dx = [2*-8*1 - 1]/2
The slope if the required tangent is -8.5
We'll differentiate the expression of the given curve, with respect to x, both sides:
d/dx[ln(x+y)] = d(x^2)/dx - dy/dx - d(89)/dx
(1/(x+y))*d(x+y)/dx = 2x - y' - 0
(1 + y')/(x+y) = 2x - y'
1 + y' = (x+y)(2x-y')
We'll remove the brackets:
1 + y' = 2x^2 - x*y' + 2xy - y*y'
We'll shift all the terms that contain y' to the left side:
y' + x*y' + y*y' = 2x^2 + 2xy - 1
We'll factorize by y':
y'*(1 + x + y) = 2x^2 + 2xy - 1
y' = (2x^2 + 2xy - 1)/(1 + x + y)
We'll calculate the slope at the point (9,-8):
y' = (162 - 144 - 1)/(1+9-8)
y' = 17/2
But y' = m
The slope of the tangent line to the given curve is m = 17/2.
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