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Just write the expansion series .i.e. expand the summation series for sinx and sin3x both up to the power of 3 of x variable. and the denominator x^3 would get cancel out. Then after put x as 0 limit ...
Find first first order derivative, dy/dx = x^n/x +nx^(n-1)*logx d2y/dx2 = (n-1)x^(n-2) + n[1/x * nx^(n-1) + logx * n(n-1)x^(n-2)] . . . . Genralizing, y(n) = n!/x somthing like that. U can also...
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TOTAL DERIVATIVES is the derivative of the function with having more than 2 variables. Just like f(x,y,t) or f(x,y,z) And here derivative of f w.r.t t always assumes that all other arguments here...
it refers to the very very small interval of that postion w.r.t itself. Limiting the change in time to 0 sec.
Rationalise it or apply H-sopital. => (x^2-4)* (√x+2 + √3x-2) / (x+2 - 3x + 2) => (x^2-4)* (√x+2 + √3x-2)/(-2x +4) =>(x-2) (x+2) (√x+2 + √3x-2)/-2(x-2) => -2(x+2) (√x+2 + √3x-2) Simply Now put x = 2...
Find out their slopes . 4y.dy/dx = 3x^2 (dy/dx)1 = 3x^2/4y............(1) and 2y.dy/dx = 32 (dy/dx)2 = 16/y...............(2) m1 * m2 = > -1 check l.hs. 3x^2/4y * 16/y = > 12x^2/y^2 put y^2 = 32x =>...
The question is incomplete, did you forget to attach the image.........................................................
The question is sin(dy/dx) = 0 If we have a look at the function , it’s not a polynomial function so it’s degree is not defined.
By definition the modulus function is defined as = -x+3 Hence derivative = -1 for all values of x less than 3 which includes -1
Inverse of f(x) is: y = ax + 1 ax = y – 1 x = (y-1)/a or y(x) = (x-1)/a then, (x-1)/a = (1+ax) solve the quadratic and fin the dicriminant and its range.
arctan((1 + (ax))^2 – 1)/ax) w.r.t 2arcsinx, Write arctan((1 + (ax))^2 – 1)/ax) in terms of sin and cos. and then differentiate it w.r.t x and out x=0.
Solve first mod, Take case x 2 is not possible according to its domain. cos(2x-4)-33/2 2 |4x-8|/x-2 becomes: cos(2x-4)-33/4 (- 4x + 8)/x-2 Now apply simply L-Hospital.put x = 2.
I answered exactly same questions for another user 3 days ago. Please check the answer for previous question under same topic.
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tHE answer should be zero itself. As no determinancy forms here. Range of cosx is [-1,1] which is a constant fixed number. Therefore power of zero as anythng constant wiould be zero.
Hello Mohan The question is not clear, could you post what needs to be proved in this question.-------------------
Here if U are using log expansion then write for cos function too. lim x-> 0 becomes: => (x(1-x^2/2) – (x – x^2/2))/x^2 => (1-x^2/2) – 1 + x/2)/x => (-x^2/2) + x/2)/x => (-x/2 + 0.5)/1 => 0.5 would...
Differentiate it first and then equate it to zero. 2cos2x – 1 = 0 cos2x = ½ 2x = pi/3 x = pi/6 or 2cos^2x – 1 = ½ cos^2x = ¾ cosx = root3/2 and also, cosx = -root3/2 x = pi/6 and x = -pi/6 check at...
put x = 2 in Lim x->2 (2/(x+2) + 1/x2-2x+4 - 24/(x3+8) =>2/4 + ¼ – 24/16 => 0.5 + 0.25 – 1.5 = > -.75 or -3/4
y = (ax+b)/(cx+d) multipy and divide by c y = (acx+bc)/c(cx+d) add and subtract ad y = (acx+ad+bc-ad)/c(cx+d) take a common from the first two terms y = a(cx+d) +bc-ad/c(cx+d) y= [a(cx+d)/c(cx+d)] +...
If the pair of state lines joining the origin to the point of intersection of the ellipse x²∕𝐲²+y²⁄𝐛²=1 and the line 𝑙𝑥+𝑚𝑦+𝑛=0 are parpandecular to each other then show that...
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