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Find the slope of the tangent line

Find dy/dx for x2y+ √2x + 5y=7
Show that the point (x,y) = (2,1) lies on the curve defined by the equation above and find the slope of the tangent line at this point. 

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What part is under the square-root?
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2 Answers

Use implicit differentiation using the chain rule to find dy/dx.  Your derivative will be in terms of x and y.  Plug in x=2 and x=1 to evaluate dy/dx.  This will give you the slope of the tangent line.
 
x2y4 + (2x + 5y)½ = 7
 
Plugging in x = 2 and y = 1, we get 7 = 7.  So, the point (2,1) lies on the curve.  
 
Implicitly differentiate the given function to obtain:
 
2xy4 + 4y3(dy/dx)(x2) + (½)(2x + 5y)(2 +5dy/dx) = 0
 
Plug in x = 2 and y = 1:
 
  4 + 16(dy/dx) + (1/6)(2 + 5dy/dx) = 0
 
  4 + 16(dy/dx) + 1/3 + (5/6)(dy/dx) = 0
 
                                   (101/6)(dy/dx) = -13/3
 
                                                dy/dx = (-13/3)(6/101) = -26/101