MAT-140 Test 2 vocab
How do you find the derivative of the sum of two functions, f+g? : The derivative of the sum of two functions, f+g, is the sum of the derivatives of each function, f'+g'.
Let H(x)= f(x) +4g(x), where the graphs of f and g are shown in the figure to the right. Find H'(3). : Use the sum and constant multiple rules to obtain the equation H'(x) = f'(x) +4g'(x). Then examine the graph to find that at x=3, the slope of f(x) is -1.5 and the slope of g(x) is 0.5. Substitute these values into the equation for H'(x) to find H'(3). H'(3) = f'(3) +4g'(3)=(-1.5) +4(0.5)=0.5
Find and simplify the derivative of the following function/f(x) = x/ x+2 : Which of the following shows how to find the derivative of f(x)?f'(x) = (x+2) (d/dx (x)) - (x) (d/dx (x+2)) / (x+2)^2
quotient rule : d/dx [f(x)/g(x)] = g(x)f'(x) - f(x) g'(x) / [g(x)]^2
Evaluate the limitlim x - 0 sin 10x / sin 11x : lim x - 0 sinx / 1 = 1lim x - 0 sin 10x/x /sin 11x/xlim x - 0 10 sin 10x /10x / sin 11x/xlim x - 0 10 sin 10x/ 10x / 11 sin 11x / 11x10/11 lim x- 0 sin 10x/ 10x / sin 11x / 11xlet t=10x in the numerator, and u = 11x in the denominator. As x - 0, t - 0, and u - 0.10/11 lim t - 0 sint/t / lim u - 0 sinu/u10/11 times 1/110/11
Use the trigonomic limits lim x - 0 sinx / x =1 and/or lim x - 0 cosx-1/x =0 to evaluate the following limit.lim x -7 sin (x-7) / x^2 -49 : Notice that the argument of the sine function is x - 7. In order to get that part of the limit to look like lim x - 0 sin x / x, use the variable transformation t= x -7Substitute x=7 into the transformation t=x-7 to determine what t approaches. t - 7 -7 or t - 0Rewrite the numerator sin (x-7) using the transformation t = x-7. sin (x-7)=sin tx^2-49=(x+7)(x-7)Notice that one of the factors is t = x-7. Use x=t +7 to transform the second factor and simplify. (x+7)(x-7) = (t+7+7)(t)=(t+14)tTherefore, the limit becomes lim t -0 sin (t) / t(t+14).Write this limit as the product of lim t -0 sint/ t and another limit, assuming that both limits exist. lim t -0 sin(t)/t(t+14) = lim t -0 sint/ t lim t -0 (1/t+14)Evaluate each limit separately. lim t -0 sint/t =1 and lim t -0 1/ t+14 = 1/14Then, multiply to find the limit.lim t -0 sint/t lim t-0 1/t+14 = 1 times 1/14 =1/14Answer = 1/14
Calculate the derivative of the following function. y = square root 14x +5 : Suppose y=f(u) is differentiable at u=g(x) and u=g(x) is differentiable at x. The composite function y =f(g(x)) is differentiable at x, and its derivative can be expressed in the two equivalent ways shown below.d/dx (f(g(x)) = f'(g(x)) times g'(x) or dy/dx = dy/du times du/dxIdentify the outer function f(u) and the inner function g(x). In this case, the best choice for these functions is f(u) = square root u and g(x) = 14x+5.To find the derivative of the outer function, f(u) = square root u, use the Power Rule (general form). If n is any real number then d/dx (x^n) = nx^(n-1) Substitute f(u) = square root u = u^1/2 into the power rule and simplifyf'(u) = 1/2 u^-1/2f'(u) = 1/2 square root uEvaluate this at g(x) = 14x+5f'(u) = 1/2 square root uf'(g(x)) = 1/2 square root 14x+5The Constant Multiple Rule states that if f is differentiable at x and c is a constant, then d/dx (cf(x)) = cf'(x).The sum Rule states that if f and g are differentiable at x, then d/dx (f(x) +g(x)) = f'(x) +g'(x).Use these rule to find the derivative of the inner function, g(x) =14x+5g(x)=14x+5g'(x) = 14Substitute f'(g(x)) = 1/2 square root 14x+5 and g'(x) =14 into the expression for d/dx (f(g(x)) shown earlierd/dx (f(g(x))) = f'(g(x)) times g'(x)=7/ square root 14x + 5Answer = 7/square root 14x + 5
Calculate the derivative of the following functiony=9 (3x^ 5 + 7) ^ -7 : dy/dx = dy/du times du/dxFirst, identify the inner function, g(x), and outer function, f(u).The inner function should be g(x) = 3x^5 +7, and the other function should be f(u) = 9u^-7.u= g(x)f(u) = 9u ^ -7f'(u) = -63u^-8g(x) = 3x^5 +7g'(x) = 15x^4Apply chain rule:dy/dx = f'(u) times g'(x)=-63u ^ -8 times 15x^4= -945x ^ 4 / (3x^5+7) ^ 8Answer = -945x ^ 4 / (3x^5 +7) ^8
Show that d/dx (ln kx) = d/dx ln x, given that x >0 and k >0 is a real number. : d/dx (ln kx) = d/dx lnud/dx ln u = 1/u d/dx ud/dx (ln kx) = 1/kx d/dx (kx)d/dx (ln kx) = 1/xCompare the result of computing d/dx (ln kx) to the derivative d/dx ln x.B. The derivatives are equal.
State the rule of differentiation for the logarithmic function f(x) = log base b x. How does it differ from the derivative formula for ln x? : Which of the following is the rule of differentiation for the logarithmic function f(x) = log base b x?B.) If b>0 and b does not equal 1, then d/dx (log base b x) = 1/xlnb for x>0, and d/dx (log base b and abolute value x) equals 1/x ln b for x does not equal 0.How does the rule of differentiation for the function f(x) = log base b x differ from the derivative for ln x?B. d/dx (log base b x) = (d/dx ln x) / ln b.
