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Derivative of Log base 4 [( x^2+1)]=y the long way?
Derivative of Log base 4 [( x^2+1)]=y the long way? Calc BC
if you change to 4^y=x^2+1 and then hy cant you NOT have the capatilized part? yln(4)=LN(X^2)+LN(1) ...why is that step not correct?
1 Answer
- JeremyLv 57 years ago
the logarithm function does not distribute over the addition like that,
log(a + b)
does not equal
log(a) + log(b)
we have:
4^y = x^2 + 1
ln(4^y) = ln(x^2 + 1)
y*ln(4) = ln(x^2 + 1)
ln(4) dy = 2x/(x^2 + 1) dx
dy/dx = 2x/(x^2 + 1) * (1/ln(4))
dy/dx = 2x/(x^2 + 1) * (1/ln(2^2))
dy/dx = 2x/(x^2 + 1) * (1/(2ln(2)))
dy/dx = x/(x^2 + 1) * (1/ln(2))
dy/dx = x/[ln(2)*(x^2 + 1)]
or, if u prefer:
dy/dx = x/ln(2^(x^2 + 1))
