Find dy/dx of the next relations
I have used Product rule of Differentiation to find out required answer.
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Find dy/dx of the next relations 16) y e'B 17) y VB 18) y [e'(B] e (Br) are cos(x) 19) y 20) y [e [5 ] 21)y [lo...
Find dy/dx of the next relations Sol: y ylx 1 1-Cx C 2) 1+cx y' (1+cx1-cx a+bx ab 3) y= In Va-bx y's a2-b'x 4) y= atan (t); x = bcor (t) 6) x+2 7) y = 2v+ 45; donde v 52., W sec X 8)4x+3 8xy+e -e+ 8cos[tan(y)] = 0 arcsec (); x = elog2 (Int) 9) y 10) y sen[tan(x )] 11) y = cos[sen' (x)] cot + 4 12) y 13) y = [sec'(secx))P 14) y [Beae...
(a) [1[*(2x*y + 4y2) dy dx (b) ["" ["(y cos(x) + 6) dy dx cos [**(buye* * *) ay ox (a) LiS.*r-* log(4) dy dx (-x log(y)) dy dx -Il
dy Find dx for the following function. y = 9 sin x + 5 cos x dy dx 1
Find a solution 6. (e* sin y + tan y)dx + (e* cos y + x(sec y)2)dy = 0.
find a solution 6. (e* sin y + tan y)dx + (e* cos y + x(sec y)2)dy = 0.
2. Solve the DE: (3x²y + cos x) dx +(x +e") dy = 0
Find dy/dx of the next relations 12)y - cor +4 13) y [sec (secx)] 14) y [Be"ae ] 15) y = [ vsec(x2 +8) arcsen(x + 8)) 12)y - cor +4 13) y [sec (secx)] 14) y [Be"ae ] 15) y = [ vsec(x2 +8) arcsen(x + 8))
4. (a Let (sin( x cos( ) dr + (x cos(x + y) - 2) dy. dz= Show that dz is an exact differential and determine the corresponding function f(x,y) Hence solve the differential equation = z sin( Cos( y) 2 x cos( y) dy 10] (b) Find the solution of the differential equation d2y dy 2 y e dx dæ2 initial conditions th that satisfi 1 (0) [15] and y(0) 0 4. (a Let (sin( x cos( ) dr...
19-20-21 19. [5pts.Each] Find the derivative of the functions: sin x a. y = b. g(x)=e* cosh(x²) 20. [5pts.] ALSO Q3 f(x)= x - cos x, 0<x< 277. Find the intervals of increasing & decreasing and intervals of concave up & concave down. Do not graph but find all relative/local maximum & minimum and inflection points if any. 21. [5pts.] Find the dimension of the rectangle of the largest area that can be inscribed in a circle of radius r....
Find the solution to the initial value problem: dy dy/dx=x^ 2√1 + x^3/1+cos y y(0)=2 the 1+x^3 is all in square root.