Asked by lana
differentiate y = ln(e^x + e^-x)
Answers
Answered by
MathMate
substitute u=e<sup>x</sup>+e<sup>-x</sup>
d(log(e^x+e^-x))/dx
=d(log(u))/dx. du/dx
= 1/u * (e<sup>x</sup>-e<sup>-x</sup>)
= (e<sup>x</sup>-e<sup>-x</sup>)/(e<sup>x</sup>+e<sup>-x</sup>)
d(log(e^x+e^-x))/dx
=d(log(u))/dx. du/dx
= 1/u * (e<sup>x</sup>-e<sup>-x</sup>)
= (e<sup>x</sup>-e<sup>-x</sup>)/(e<sup>x</sup>+e<sup>-x</sup>)
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