what is dy/dx if y=(3x^4+4x^3+1)(3x^2+2x^3-1)?
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Tushar Chandra
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We have y = (3x^4+4x^3+1)(3x^2+2x^3-1)
To find the derivative dy/dx we use the product rule.
y' = (3x^4+4x^3+1)'(3x^2+2x^3-1) + (3x^4+4x^3+1)(3x^2+2x^3-1)'
y' = (12x^3+12x^2)'(3x^2+2x^3-1) + (3x^4+4x^3+1)(6x+6x^2)'
=> y' =12( 3x^5 + 2x^6 - x^3 + 3x^4 + 2x^5 - x^2) + 6(3x^5 + 4x^4 + x +3x^6 + 4x^5 + x^2)
=> ( 36x^5 + 24x^6 - 12x^3 + 36x^4 + 24x^5 - 12x^2) + (18x^5 + 24x^4 + 6x +18x^6 + 24x^5 + 6x^2)
=> ( 42x^6 +102x^5 + 60x^4 - 12x^3 - 6x^2 + 6x
The required result is ( 42x^6 + 102x^5 + 60x^4 - 12x^3 - 6x^2 + 6x
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