Exam 1

  1. Question

    Let

    Select each of the true statements from the following:


    1. For k(x)k(x) the outside function is f(x)=3x2f(x)=3x^2 and the argument is g(x)=sin(x)g(x)=\sin(x)
    2. For v(x)v(x) the outside function is f(x)=exf(x)=e^x and the argument is g(x)=6.73xg(x)=6.73x
    3. For j(x)j(x) the outside function is f(x)=ln(x)f(x)=\ln(x) and the argument is g(x)=12x+11g(x)=12x+11
    4. j(x)=112x+11j^'(x) = \frac{1}{12x+11}
    5. v(x)=6.73xe6.73x1v^'(x) = 6.73xe^{6.73x-1}
    6. k(x)=cos(6x)k^'(x) = \cos(6x)
    7. All of the above statements are false.

    Solution

    1. False. The composition of f(x)=3x2f(x)=3x^2 and g(x)=sin(x)g(x)=\sin(x) is f(g(x))=f(sin(x))=3(sin(x))2k(x) f(g(x))=f(\sin(x))=3(\sin(x))^2 \ne k(x)
    2. True. For v(x)=e6.73xv(x)=e^{6.73 \cdot x} the outside function is f(x)=exf(x)=e^x and the argument (or inside function) is g(x)=6.73xg(x)=6.73x. The composition of f(x)f(x) and g(x)g(x) is f(g(x))=f(6.73x)=e6.73x=v(x) f(g(x))=f(6.73x)=e^{6.73x}=v(x)
    3. True. For j(x)=ln(12x+11)j(x)=\ln(12x+11) the outside function is f(x)=ln(x)f(x)=\ln(x) and the argument (or inside function) is g(x)=12x+11g(x)=12x+11. The composition of f(x)f(x) and g(x)g(x) is f(g(x))=f(12x+11)=ln(12x+11)=j(x) f(g(x))=f(12x+11)=\ln(12x+11)=j(x)
    4. False. What function(s) could have derivative 112x+11\frac{1}{12x+11}? Since the derivative of ln(g(x))\ln(g(x)) with respect to xx is 1g(x)g(x)\frac{1}{g(x)}\cdot g^'(x) you can check that the derivative of 112ln(12x+11)\frac{1}{12}\ln(12x+11) is an answer to this question.
    5. False. What function(s) could have derivative 6.73xe6.73x16.73xe^{6.73x-1}? This question is beyond the scope of this course but the process used to find an answer involves the product rule. Incorrect use of the power rule may be the reason someone would think ddxe6.73x=6.73xe6.73x1\frac{d}{dx}e^{6.73x} = 6.73xe^{6.73x-1}
    6. False. The chain rule states ddxsin(3x2)=k(x)=f(g(x))g(x)\frac{d}{dx}\sin(3x^2)=k^'(x)=f^'(g(x))g^'(x) where the outside function is f(x)=sin(x)f(x)=\sin(x) and the argument (or inside function) is g(x)=3x2g(x)=3x^2. Since f(x)=cos(x)f^'(x)=\cos(x) (not cos(x)-\cos(x)) and g(x)=6xg^'(x)=6x k(x)=cos(g(x))g(x)=cos(3x26xk^'(x)=\cos(g(x))\cdot g^'(x)=\cos(3x^2\cdot 6x
    7. False.