Differentiation

Calculus Level 1

If y = ( x − 1 ) ( 2 x 2 + 3 ) y=(x-1)(2x^2+3) , find d y d x \dfrac {dy}{dx} .

6 x 3 + 4 x + 3 6x^3 +4x +3 6 x 2 − 4 x + 3 6x^2 - 4x +3 6 x 2 + 4 x − 3 6x^2+ 4x -3 6 x 2 − 6 x + 3 6x^2-6x +3

Saraswati Sharma by Saraswati Sharma , 23, India

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1 solution

Viki Zeta
Oct 13, 2016

d d x ( x − 1 ) ( 2 x 2 + 3 ) = d d x u v ; u = ( x − 1 ) , v = ( 2 x 2 + 3 ) = u v ′ + v u ′ = ( x − 1 ) d d x ( 2 x 2 + 3 ) + ( 2 x 2 + 3 ) d d x ( x − 1 ) = ( x − 1 ) ( 4 x ) + ( 2 x 2 + 3 ) = 4 x 2 − 4 x + 2 x 2 + 3 = 6 x 2 − 4 x + 3 \dfrac{d}{dx} (x-1)(2x^2+3) = \dfrac{d}{dx} ~ uv; ~~ u = (x-1), \ v = (2x^2+3) \\ = uv' + vu' = (x-1)\dfrac{d}{dx}(2x^2+3) + (2x^2+3)\dfrac{d}{dx}(x-1) \\ = (x-1)(4x) + (2x^2+3) \\ = 4x^2 - 4x + 2x^2 + 3 \\ = 6x^2 - 4x + 3

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Vicky Vignesh Actually the question was to find dy/dt . Thats why I posted this problem. Initially I also thought that instead of dy/dt there should be dy/dx . But this problem is from chain rule p-set so probability of doing this problem by chain rule increases. (by substitution)

Saraswati Sharma - 4 years, 8 months ago

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Product rule is pretty easier than chain rule in this case.

Viki Zeta - 4 years, 8 months ago

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Conclusion - There should be dy/dx in p-sets rather than dy/dt which is creating a lot of confusion . There r straight 8 questions in my p-set with this same mistake .

Saraswati Sharma - 4 years, 8 months ago

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