How to find dy/dx?? calculus 1?

Please can you help me to solve these questions quickly ??

Find dy/dx of:

A))) y =(1/x)*e^y + e^(1+y)^2 ,

B))) ln y + (x^2)*(y)^1/2- sinh(xy) = e^2,

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  • 6 years ago
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    No, not dy/dx "of". It would be dy/dx "if" - since what follows is an equation. Oh well, here are the answers...!

    (A) y = (1/x) e^y + e^[(1+y)^2]

    dy/dx = (-1/x^2) e^y + (1/x) e^y dy/dx + e^[(1+y)^2] (2y+2) dy/dx

    (1/x^2) e^y = dy/dx { (1/x) e^y + e^[(1+y)^2] (2y+2) - 1 }

    e^y = x dy/dx { e^y + xe^[(1+y)^2] (2y+2) - x }

    dy/dx = e^y / { xe^y + x^2 e^[(1+y)^2] (2y+2) - x^2 }

    (B) ln(y) + (x^2)*[y^(1/2)] - sinh(xy) = e^2

    (1/y) dy/dx + 2x [y^(1/2)] + (x^2)(1/2)[y^(-1/2)] dy/dx

    - cosh(xy) [x dy/dx + y] = 0

    Multiplying by y^(3/2) gives

    y^(1/2) dy/dx + 2xy^2 + (1/2)yx^2 (dy/dx)

    - y^(3/2) cosh(xy) [x dy/dx + y] = 0

    Rearrange terms

    (dy/dx) * [y^(1/2) + (1/2)yx^2 - xy^(3/2) cosh(xy)]

    = y^(5/2) cosh(xy) - 2xy^2

    Final step is obvious, just divide RHS by the bracketed coefficient from the LHS.

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