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Find the derivative dy/dx using chain rule for y = x^3 (2x -5)^4
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'(x^3+x^2+x+1) dy/dx=2x^2+x ' y=1 when x=0 || '(x^3+x^2+x+1) dy/dx=2x^2+x'
Differentiate y=(x+1)^3(x+2)^4(x+3)^5 / y=(2-x)^3(3+2x)^5,find dy/dx / Differentiation Trick for JEE
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If f(x) = 2x^2 - 3x + 5, find f(x+1).
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