Recent content by JackieAnne

  1. J

    Solving Concave Functions: Intervals & Inflection Points

    okay, so I think I got the first derivative: 6x^5*ln(x) + x^6*(1/x) so then would the second derivative be: 30x^4*ln(x) + x^6*(-1/x^2)
  2. J

    Solving Concave Functions: Intervals & Inflection Points

    Let f(x)=x^6ln(x) . Find (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. (a) f is increasing on the...
  3. J

    How Do You Correctly Solve the Limit of (6π−6x)tan(x/2) as x Approaches π?

    Find the limit. limx->pi- (6pi−6x)tan(x/2) = lim(x→π) [ (6π - 6x) tan(x/2) ] = 6 lim(x→π) [ (π - x) tan(x/2) ] = 6 lim(x→π) [ (π - x) / cot(x/2) ] apply l'Hôpital's Rule since it's in the form of 0/0, at x = π = 6lim(x→π) [ d/dx (π - x) / d/dx (cot(x/2)) ] = 6lim(x→π) [ -1 / ((1/2)...
  4. J

    So the correct answer isdy/dx = (-sin4x - 4siny) / (4y + 4cos4x)

    Use implicit differentiation to find dy/dx. y is a differentiable function of x 2y^2+4xsiny = cos4x Here is what I did: 4y*dy/dx + 4siny+ 1/cosy*dy/dx = -sin4x + 4 4y*dy/dx + 1/cosy*dy/dx = -sin4x - 4siny + 4 dy/dx(4y + 1/cosy) = -sin4x - 4siny + 4 dy/dx = (-sin4x - 4siny + 4) /...
  5. J

    Finding the Derivative of a Square Root Function

    Hi so I am new to calculus and need help with differentiation The question is: Differentiate the following function: sqrt(x^2-8x+20) So far I have gotten =(x^2-8x+20)^(1/2) =(1/2)(x^2-8x+20)^(-1/2) * 2x-8 I am unsure where to go from here I know we are supposed to use the chain...
  6. J

    Derivatives, Sin and Cos, Rate of Change, Tangent Lines

    Hi, I am in calculus and am having major struggles. If someone could provide a walk through on how to answer these questions, that would be fantastic. Cheers! Let f(x)=−3x+6 if x<-3 = 15 if x > -3 Find the average rate of change of f(x) on the interval −5<x<5 . The average rate of...
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