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...
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...
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