If y = (x – 1)(2x + 3)(sec x); evaluate y'. Strategy # 1 - multiply out 1st, then differentiate: y = (2x^2 + x – 3)(sec x) = 2x^2(sec x) + x(sec x) – 3(sec x) y' = 2x^2(sec x)(tan x) + 4x(sec x) + x(sec x)(tan x) + (sec x) – 3(sec x)(tan x). Strategy # 2 - use the product rule, then segregate and recombine like terms: y' = (x – 1)(2x + 3)(sec x)(tan x) + 2(x – 1)(sec x) + (2x + 3)(sec x) = (2x^2 + x – 3)(sec x)(tan x) + 2x(sec x) – 2(sec x) + 2x(sec x) + 3(sec x) = 2x^2(sec x)(tan x) + x(sec x)(tan x) – 3(sec x)(tan x) + 2x(sec x) – 2(sec x) + 2x(sec x) + 3(sec x) = 2x^2(sec x)(tan x) + 4x(sec x) + x(sec x)(tan x) + (sec x) – 3(sec x)(tan x). Strategy # 3 - take the natural log of both sides, then differentiate: ln y = ln(x – 1) + ln(2x + 3) + ln(sec x) y'/y = [1/(x – 1)] + [2/(2x + 3)] + [(sec x)(tan x)/(sec x)] = [1/(x... read more
If the function “cos sqrt(x) dx” has to be integrated, would you consider using the substitution method? Let w = sqrt(x), dw = dx/2sqrt(x), or 2dw = dx/sqrt(x) . . . Are we stuck? Typically, we employ the substitution method for integration when we can identify a function and its derivative within the integrand. For example, in integral[cos x sin x dx], we recognize the function “sin x” and its derivative “cos x dx”, so the substitution methodology here is straightforward. We let u = sin x and du = cos x dx. However, for integral[cos sqrt(x) dx] we must be creative and remember that w = sqrt(x), then 2dw = dx/sqrt(x) = dx/w, or 2wdw = dx.* Now, integral[cos sqrt(x) dx] = integral[2w cos w dw] = 2 integral[w cos w dw.] This integral is now fit for integration-by-parts using the “Liate” rule where we let “u” equal the algebraic vs the trigonometric function. Then we conclude by reversing the substitution. Let u = w, dv = cos w dw, so du = dw and v = sin... read more
The tool of mathematics is easily considered the most exact of the sciences. However, if these usually very precise answers to homework and exam problems were cities or towns on a map, then part of math's beauty is that several roads can lead into it. We have perhaps seen or worked word problems which could be solved using more than one branch of math, for example, calculating the required dimensions of a tunnel entrance utilizing either geometry or trigonometry. Calculus is understandably difficult for many, if not most, of us, but its arsenal of differentiation or integration techniques allow it to be quite flexible when its applied to homework assignments or exam problems. For instance, differentiating a polynomial divided by another polynomial can achieved (after any simplifications) using either the quotient rule or the product rule (from which the former is derived anyway). If differential calculus is primarily protocol, then integral calculus is significantly... read more
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