If f(x) = e^[3ln(x^2)], then f'(x) = ?
2 answers:
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Answer: at the values where cos(x) = 0
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.