For |x| <1, the derivative of y=ln√1-x² is?
Let's solve the derivative and see what we get. y = ln (sqrt(1 - x^2)) For one thing, you can use log properties to transform this into something simpler. y = ln [ (1 - x^2)^(1/2) ] Using the log property log[base b](a^c) = c*log[base b])(a), y = (1/2) ln (1 - x^2) Factoring the inside of the log, y = (1/2) ln [(1 - x)(1 + x)] Applying the log property log[base b](ac) = log[base b](a) + log[base b](c), we get: y = (1/2) [ln (1 - x) + ln(1 + x)] Keep in mind at that this point we haven't even taken the derivative yet; only simplified its form. Now, taking the derivative, we get y = (1/2) [ {1/(1 - x)}(-1) + 1/(1+x)} y = (1/2) [1/(1 + x) - 1/(1 - x)] We can change the form of this, to y = (1/2) [ {1 - x} - {1 + x} ] / [(1 - x)(1 + x)] y = (1/2) [-2x]/[(1 - x^2)] Merging this into one term, we finally get y = -x/(1 - x^2) But we can factor out a -1 in the denominator. y = -x / [-1(x^2 - 1)] And now we can cancel the negative signs, y = x / (x^2 - 1) For that reason, the answer is (b).
Find the derivative of the function?
We can estimate the derivative f'(2.1) by secant slope formula. m = (f(b) - f(a))/(b - a) at [a,b] Take: a = 2 b = 2.2 So we have.. m = (f(2.2) - f(2))/(2.2 - 2) = (1.9^(1.2 * 2.2) - 1.9^(1.2 * 2))/0.2 = 3.88603 ≈ 3.89 Check by differentiation! Set both sides by ln to get.. ln(f(x)) = 1.2xln(1.9) Then, by implicit differentiation, noting that d/dx ln(f(x)) = f'(x)/f(x).. f'(x)/f(x) = 1.2ln(1.9) [Also noting that 1.2ln(1.9) altogether is the constant. Differentiate x.] f'(x) = f(x) * 1.2ln(1.9) [Multiply both sides by f(x)] f'(x) = 1.9^(1.2x) * 1.2ln(1.9) Hence, f'(2.1) ≈ 3.88. The previous value is close to that value by 0.01 I hope this helps!
Calculus..find the derivative of the function?
number 45. (x^2 - 3x + 2) vdu - udv y= -------------------- ; from the law of calculus d[u/v] = ------------- (x^7 - 2) v^2 dy = { [ (x^7 - 2)(2x - 3) - (x^2 - 3x + 2)(7x^6) ] / (x^7 - 2)^2 } dx dy [ (2x^8 - 3x^7 - 4x + 6) - (7x^8 - 21x^7 +14x^6) ] ---- = ----------------------------------------... dx (x^7 - 2)^2 dy [-5x^8 + 18x^7 + 14x^6 - 4x + 6] answer is: ---- = ----------------------------------------... dx (x^7 - 2)^2 for number 46: d(y) = d(4x^3 - 7x^2 +3) ; using simple derivation and for U^n = nU^n-1 dy ---- = (12x^2 - 14x) dx the second derivative is: d²y ----- = 24x - 14 <------ this is the answer for number 46! dx²
How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.
How to solve using logarithmic differentiation step by step?
y = x^(6x) Start off by taking the natural log of both sides in order to bring down the 6x exponent: ln(y) = (6x) ln(x) Now use implicit differentiation: (1/y)y' = (6x)(1/x) + (lnx)(6) This equals: (1/y)y' = 6 + 6ln(x) Multiply both sides by y: y ' = y[6 + 6ln(x)] Now remember from the beginning that y = x^(6x) so you can substitute that in for the y: y ' = (x^(6x))[6 + 6ln(x)] You can factor out a 6 from the 6 + 6ln(x) to give you: y ' = (6x^(6x)) [1 + ln(x)] <===Answer