write the equation of the tangent line through (1,2) for sqrtx/y^2+4+15/8=2x

Thank you for the opportunity to help you with your question!

The answer ends up being:

y=(1985-63x)/961

:) The full derivation requires more time than this question allows to write out. Let me know if there's anything more I can help with.

can you tell me where I went wrong see below:

sqrtx/ y ^2+4 +15/8 = 2x

sqrtx/ y ^2+4 +1.87 = 2x

sqrtx/ y ^2= 2x + 5.87

sqrt x/ y ^2=2x + 5.87

y^2 = sqrtx /(2x + 5.87)

y = sqrt [sqrt(x )/ (2x + 5.87)]

by adding given point

2= sqrt [sqrt(1 )/ (2(1) + 5.87)]+c

2= sqrt [1 / 7.87]+c

5.6= c

equation of tangent

Y= M x+C

y= 4 x + 5.6

Arg! I had a whole response written up but clicked the back button.

Don't simplify the 15/8. With calculus, decimals are a headache to deal with. When and where possible, just keep things as fractions.

Simplify the expression by clearing denominators, so multiply every term by 8y^2:

(8y^2)(sqrtx/y^2) + (8y^2)4 + (8y^2)(15/8) = 2x(8y^2)

8sqrtx + 32y^2 + 15y^2 = 16xy^2

No more denominators! Which means we can avoid the annoying rules about deriving with denominators. Now we just need to combine like terms:

8sqrtx + 47y^2 = 16xy^2

And then isolate y:

8sqrtx = 16xy^2 - 47y^2

8sqrtx = y^2(16x - 47)

y^2 = 8sqrtx/(16x - 47)

And then the first derivative:

2y=(4(16x-47)/sqrt(x)-128sqrt(x))/(16x-47)^2

Solve for point (1,2) to find the slope at that point, which ends up being:

m = -63/961

And that's it! That's the hard point. :) The rest is just finding b with the slope of a line equation.

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