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I would really appreciate any help with this one I am stuck between two answers for it...

2007-06-02 05:19:48 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

Ok, remember :

dcos(hx) / dx = sin(hx)

dy / dx, when : y = cosh(bx^2)

dy / dx = 2*b*x*sinh(bx^2)

Hope that helps

2007-06-02 05:24:33 · answer #1 · answered by anakin_louix 6 · 0 0

Is that cosh(bx^2)?
dy/dx = 2*bx * sinh(bx^2)

2007-06-02 12:32:55 · answer #2 · answered by Dr D 7 · 0 0

Let z = bx^2
dz/dx = 2bx

y = cosh(bx^2)
y = cosh(z)
dydz = sinh(z)

dy/dx = (dy/dz).(dz/dx)
dy/dx = [sinh(z)][2bx]
dy/dx = 2bxsinh(bx^2)

2007-06-02 17:15:42 · answer #3 · answered by Kemmy 6 · 0 0

y=cosh(bx^2)
dy/dx=2bx sinh(bx^2) answer

2007-06-02 12:24:34 · answer #4 · answered by Anonymous · 0 0

let u = bx²
du/dx = 2 b x
y = cosh u
dy/du = sinh u
dy/dx = (du/dx).(dy/du)
dy/dx = (2bx).sinh (bx²)

2007-06-02 13:51:32 · answer #5 · answered by Como 7 · 0 0

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