How to use the definition to find the derivative of LNX,
derivatives
=[ln(x+h)-lnx]/h
= ln[(x+h)/x]/h
=1 / XLN (1 + H / x) / h / x H tends to 0
=1/X
It's over
RELATED INFORMATIONS
- 1. The derivative of X + LNX
- 2. Judge the number of real number solutions of the equation f (x) = LNX - (2 / x) Judge the number of real number solutions of the equation f (x) = LNX - (2 / x) Judge the number of real number solutions of the equation f (x) = LNX - (2 / x) = 0
- 3. Solve the equation LNX = e + 1-x
- 4. 1. The number of solutions of the equation x ^ 2-x = LNX is 2. Given that the function f (x) = 4 ^ x + k * 2 ^ x + 1 has only one zero point, then the zero point is 1. The number of solutions of the equation x ^ 2-x = LNX is 2. Given that the function f (x) = 4 ^ x + k * (2 ^ x) + 1 has only one zero point, then the zero point is
- 5. If f (x) = | LNX | has three intersections with the function y = KX in the interval (0,3), then the value range of K is
- 6. If the intersection of the line y = KX + 2K + 1 and the line y = − 12x + 2 is in the first quadrant, then the value range of K is () A. −12<k<12B. −16<k<12C. k>12D. k>−12
- 7. Given that the two intersections of two curves y = KX + 1 and x ^ 2-y-8 = 0 are symmetric about the Y axis, then the coordinates of the two intersections are
- 8. It is known that hyperbola C1 and hyperbola C2: y ^ 2 / 4-x ^ 2 / 9 = 1 have the same asymptote And through the point m (9 / 2, - 1), the standard equation of hyperbola C1 is obtained
- 9. Given the hyperbola c1:2x ^ 2-y ^ 2 = 1, let the ellipse c2:4x ^ 2 + y ^ 2 = 1, if M and N are the moving points on C1 and C2 respectively, and OM is perpendicular to on, we prove that: The distance from O to the straight line Mn is a fixed value
- 10. As shown in the figure, the left and right fixed points of ellipse C1: x ^ 2 / 4 + y ^ 2 / 3 = 1 are a, B and P respectively, which are a point on the right branch (above the X axis) of hyperbola C2: x ^ 2 / 4-y ^ 2 / 3 = 1 Connecting AP is called C1 to C, connecting Pb and extending the intersection C1 to D, and the areas of △ ACD and △ PCD are equal. Calculate the slope of the straight line PD and the inclination angle of the straight line CD
- 11. It is known that the right quasilinear of two hyperbolas is x = 4, the right focus is f (10.0), and the eccentricity is e = 2?
- 12. Given that the eccentricity of hyperbola is 2, the focus is (4,0), (- 4,0), then what is the hyperbolic equation? Help me if you know
- 13. The standard hyperbolic equation with focus on the x-axis, imaginary axis length of 12 and eccentricity of 54 is () A. x264−y2144=1B. x236−y264=1C. y264−x216=1D. x264−y236=1
- 14. Let the eccentricity of hyperbola be √ 5 / 2 and have a common focus with the square of ellipse x2 / 9 + the square of Y2 / 4 = 1
- 15. It is known that F1F2 is the left and right focus of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0), the line passing through point F1 and perpendicular to the real axis and the two asymptotes of hyperbola If the coordinate origin o is just the vertical center of △ abf2 (the intersection of the three high lines of the triangle), then the eccentricity of the hyperbola is
- 16. Given that the right focus of the hyperbola x2a2-y25 = 1 is (3,0), then the eccentricity of the hyperbola is equal to () A. 31414B. 324C. 32D. 43
- 17. Let the hyperbola x 2 / 4-y 2 / 9 = 1, f 1F 2 be the two focal points. The point m is on the hyperbola (1) If ∠ f1mf2 = 90 °, calculate the area of △ f1mf2 (2) If ∠ f1mf2 = 60 °, what is the area of △ f1mf2? If ∠ f1mf2 = 120 °, what is the area of △ f1mf2?
- 18. The left and right focal points of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0) are F1 and F2 respectively. The segment F1F2 is divided into two segments of 3:2 by the point (B / 2,0), which is the eccentricity of hyperbola
- 19. Hyperbola has the same focus as ellipse x square + 2Y square = 20, its asymptote is 3x-y = 0, then hyperbolic equation is?
- 20. It is known that the hyperbolic equation is x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > O, b > o), and the distance from a focus of hyperbola to an asymptote is √ 5C / 3 to calculate the eccentricity The asymptote equation of hyperbola is y = (± B / a) x, and an asymptote is y = (B / a) x or BX ay = 0 A focus of hyperbola f (C, 0), a ^ 2 + B ^ 2 = C ^ 2 The distance from focus f to an asymptote is: |bc-a*0|/c=b b=√5c/3,9b^2=5c^2 From: A ^ 2 + B ^ 2 = C ^ 2 I know the answer to this question, and then I want to ask why a ^ 2 = 4 / 9C ^ 2