differentiate x cube + 2 X square + 5 x + 1 with respect to X
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= x^3+2x^2+5x+1
= d/dx [x^3+2x^2+5x+1]
= d/dx x^3 + d/dx 2x^2 + d/dx 5x +d/dx 1
= 3x^2 + 2x + 5 + 0 (d/dx x ^n = nx^(n-1))
= 3x^2+2x^3+5
= d/dx [x^3+2x^2+5x+1]
= d/dx x^3 + d/dx 2x^2 + d/dx 5x +d/dx 1
= 3x^2 + 2x + 5 + 0 (d/dx x ^n = nx^(n-1))
= 3x^2+2x^3+5
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