Stuck on a few P3 Differentiation Q's. Watch

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Slice'N'Dice
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#1
Report Thread starter 14 years ago
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I'm getting there! Just been put off track by these Q's;

Differentiate the following functions, express answers as simply as possible;

1.) y = x√(4x+1)

2.) y = x/√(4x+3)

3.) y = (2x²-3)√(2x+1)

4.) y = 3x/√(x²+4x)


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Charlottie
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Remember that the square root is also power 1/2 and look at the quotient rule.
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mr_tomus
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or product rule (which is easier) and write the quotients as products.
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Nima
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(Original post by Slice'N'Dice)
Differentiate the following functions, express answers as simply as possible;
1.) y = x√(4x+1)
2.) y = x/√(4x+3)
3.) y = (2x²-3)√(2x+1)
4.) y = 3x/√(x²+4x)
1.) Let u = x ---> du/dx = 1
Let v = (4x + 1)^(1/2) ---> dv/dx = (1/2 . 4)(4x + 1)^(-1/2) = 2/[(4x + 1)^(1/2)]
Hence: dy/dx = v(du/dx) + u(dv/dx) = (4x + 1)^(1/2) + 2x/[(4x + 1)^(1/2)] = [(4x + 1) + 2x]/[(4x + 1)^(1/2)]
---> dy/dx = (6x + 1)/[(4x + 1)^(1/2)]

2.) y = x/[(4x+3)^(1/2)] = x(4x + 3)^(-1/2)
Let u = x ---> du/dx = 1
Let v = (4x + 3)^(-1/2) ---> dv/dx = (-1/2 . 4)(4x + 3)^(-3/2) = -2/[(4x + 3)^(3/2)]
Hence: dy/dx = v(du/dx) + u(dv/dx) = 1/[(4x + 3)^(1/2)] - 2x/[(4x + 3)^(3/2)] = [(4x + 3) - 2x]/[(4x + 3)^(3/2)]
---> dy/dx = (2x + 3)/[(4x + 3)^(3/2)]

3.) Let u = 2x^2 - 3 ---> du/dx = 4x
Let v = (2x + 1)^(1/2) ---> dv/dx = (1/2 . 2)(2x + 1)^(-1/2) = (2x + 1)^(-1/2)
Hence: y = v(du/dx) + u(dv/dx) = 4x(2x+1)^(1/2) + (2x^2 - 3)(2x + 1)^(-1/2) = 4x(2x+1)^(1/2) + (2x^2 - 3)/[(2x+1)^(1/2)] = [4x(2x + 1) + (2x^2 - 3)]/[(2x+1)^(1/2)] = [8x^2 + 4x + 2x^2 - 3]/[(2x+1)^(1/2)]
---> dy/dx = (10x^2 + 4x - 3)/[(2x+1)^(1/2)]

4.) y = 3x(x^2 + 4x)^(-1/2)
Let u = 3x ---> du/dx = 3
Let v = (x^2 + 4x)^(-1/2) ---> dv/dx = (-1/2 . (2x + 4))(x^2 + 4x)^(-3/2) = -(x + 2)(x^2 + 4x)^(-3/2)
Hence: dy/dx = v(du/dx) + u(dv/dx) = 3/[(x^2 + 4x)^(1/2)] - [(3x)(x + 2)]/[(x^2 + 4x)^(3/2)] = [3(x^2 + 4x) - 3x(x + 2)]/[(x^2 + 4x)^(3/2)]
---> dy/dx = 6x/[(x^2 + 4x)^(3/2)]
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