Differentiation with respect to x
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y=e^[log(x+√(x^2+a^2))]
Now taking log both sides we get,
logy=log(x+√(x^2+a^2)loge
logy=log(x+√x^2+a^2)
(since log e=1)
Now solve it ....
Now taking log both sides we get,
logy=log(x+√(x^2+a^2)loge
logy=log(x+√x^2+a^2)
(since log e=1)
Now solve it ....
neerajvermag11:
Thanks
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