course MTH272 cN}܊фjhۿeKassignment #003
......!!!!!!!!...................................
21:56:13 4.5.10 (was 4.4.10) find the derivative of ln(1-x)^(1/3)
......!!!!!!!!...................................
RESPONSE --> ln(1-x)^(1/3) 1/3*ln(1-x) 1/3 * (1/1-x) 1/(3-3x) confidence assessment:
.................................................
......!!!!!!!!...................................
22:00:29 The function is of the form ln(u), so the derivative is 1/u * u', or ln(u) * du/dx. The function u is (1-x)^(3/2). The derivative of this function is u' = du/dx = -1 * 3/2 * (1-x)^(1/2) = -3/2 (1-x)^(1/2). Thus the derivative of the original function is 1/u du/dx = 1 / [(1-x)^(3/2) ] * [-3/2 (1-x)^(1/2)] = -3/2 (1-x)^(1/2) (1-x)^(-3/2) = -3/2 (1-x)^-1 = -3 / [ 2 (1-x) ] ALTERNATIVE SOLUTION: Note that ln(1-x)^(1/3) = 1/3 ln(1-x) The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x). The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].**
......!!!!!!!!...................................
RESPONSE --> looks like i followed the alternative solution using the properties of logarithms.was it incorrect to distribute the denominator?
.................................................
......!!!!!!!!...................................
22:07:32 4.5.25 (was 4.4.24) find the derivative of ln( (e^x + e^-x) / 2)
......!!!!!!!!...................................
RESPONSE --> ln(z) z'= [e^x - e^(-x)] [(e^x - e^(-x))/4]*[2/(e^x+ e^(-x)] these would cancel out to= -1 (if my algebra is right) confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:08:21 the derivative of ln(u) is 1/u du/dx; u = (e^x + e^-x)/2 so du/dx = (e^x - e^-x) / 2. The term - e^(-x) came from applying the chain rule to e^-x. The derivative of ln( (e^x + e^-x) / 2) is therefore [(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x). This expression does not simplify, though it can be expressed in various forms (e.g., (1 - e^-(2x) ) / ( 1 + e^-(2x) ), obtained by dividing both numerator and denominator by e^x.). ALTERNATIVE SOLUTION: ln( (e^x + e^-x) / 2) = ln( (e^x + e^-x) ) - ln(2). the derivative of e^(-x) is - e^(-x) and ln(2) is constant so its derivative is zero. So you get y ' = (e^x - e^-x)/(e^x + e^-x). **
......!!!!!!!!...................................
RESPONSE --> not sure where my mistake is exactly self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:14:47 4.5.30 (was 4.4.30) write log{base 3}(x) in exp form
......!!!!!!!!...................................
RESPONSE --> logax=b if and only if a^b=x so- 3^y=x confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:15:03 the exponential form of y = log{base 3}(x) is x = 3^y, which I think was the question -- you can check me on that and let me know if I'm wrong **
......!!!!!!!!...................................
RESPONSE --> u got it self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:27:04 Extra Problem (was 4.4.50) Find the equation of the line tangent to the graph of 25^(2x^2) at (-1/2,5)
......!!!!!!!!...................................
RESPONSE --> i tried using the chain rule multiple times, and kept coming up with 200x^(2x^2+1) i know thats not right though- confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:27:48 Write 25^u where u = 2x^2. So du/dx = 4x. The derivative of a^x is a^x * ln(a). So the derivative of 25^u with respect to x is du / dx * ln(25) * 25^u = 4x ln(25) * 25^u = 4x ln(25) * 25^ (2 x^2). Evaluating this for x = -1/2 you get 4 * (-1/2) ln(25) * 25^(2 * (-1/2)^2 ) = -2 ln(25) * 25^(1/2) = -2 ln(25) * 5 = -10 ln(25) = -20 ln(5) = -32.189 approx. So the tangent line is a straight line thru (-1/2, 5) and having slope -20 ln(5). The equation of a straight line with slope m passing thru (x1, y1) is (y - y1) = m ( x - x1) so the slope of the tangent line must be y - 5 =-20 ln(5) ( x - (-1/2) ) or y - 5 = -20 ln(5) x - 10 ln(5). Solving for y we get y = -20 ln(5) x - 10 ln(5). A decimal approximation is y = -32.189x - 11.095 ALTERNATIVE SOLUTION: A straight line has form y - y1 = m ( x - x1), where m is the slope of the graph at the point, which is the value of the derivative of the function at the point. So you have to find the derivative of 25^(2x^2) then evaluate it at x = -1/2. The derivative of a^x is ln(a) * a^x. The derivative of 25^z would therefore be ln(25) * 25^z. The derivative of 25^(2 x^2) would be found by the chain rule with f(z) = 25^z and g(x) = 2 x^2. The result is g ' (x) * f ' (g(x)) = 4 x * ln(25) * 25^(2x^2). Evaluating at x = -1/2 we get -2 ln(25) * 25^(1/2) = -10 ln(25). Now we use the ponit-slope form of the equation of a straight line to get (y - 5) = -10 ln(25) * (x - (-1/2) ) and simplify. **
......!!!!!!!!...................................
RESPONSE --> where does the ln come from? self critique assessment: 2
.................................................