Hi, please help solve the equation f'(x) = 0 f(x)= (ln x +2x) ^ 1/3 thanks
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Originally Posted by thelazyman Hi, please help solve the equation f'(x) = 0 f(x)= (ln x +2x) ^ 1/3 thanks Given the function, Then, by the Generalized Exponent Rule, But Thus, thus, because is positive. Thus, This is my 2 Post!
Hi, test is tommorow need help with this question For each of the following, find the slope of the tangent. For the curve defined by xe^y + ylnx = 2 at (1, ln2). Thanks
Hi, I understand how you got the -1/2, but what happens to the 1/3 and (lnx + 2x)^ -2/3, do they cancel out and why>?
Originally Posted by thelazyman Hi, I understand how you got the -1/2, but what happens to the 1/3 and (lnx + 2x)^ -2/3, do they cancel out and why>? You could think of it like that. See the problem is that is never zero thus you can divide both sides by this.
Originally Posted by thelazyman Hi, test is tommorow need help with this question For each of the following, find the slope of the tangent. For the curve defined by xe^y + ylnx = 2 at (1, ln2). Thanks Take the derivative of both sides implicitly, Remember to use the product rule Simplify, Reorganize, Now factor, Remember the problem says that point Substitute those values for Thus, (remember ) Thus,
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