what dy/dx for this equation: x=e^x - (x^2+1)ln(x^2+1)?

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what dy/dx for this equation: x=e^x - (x^2+1)ln(x^2+1)?

2007-03-09 17:43:44 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

3 answers

dx/dy= (e^x) - 2x ln(x^2+1) - 2x

2007-03-09 17:48:15 · answer #1 · answered by A New Life 3 · 0⤊ 0⤋

Do you mean y = e^x - (x^2+1)*ln(x^2+1)? Then by application of the product rule and the chain rule you get:

e^x - 2x*ln(x^2+1) - (x^2+1)*[1/(x^2+1)]*2x

e^x - 2x*[ln(x^2+1) + 1]

2007-03-10 01:50:38 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋

dy/dx = -2 * x * ln(x^2 + 1) + e^x - 2x

2007-03-10 01:50:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋