EE1. Solutions of Problems 4. : a) f(x) = x 2 +x. = (x+ǫ)2 +(x+ǫ) (x 2 +x) ǫ

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1 EE Solutions of Problems 4 ) Differentiation from first principles: f (x) = lim f(x+) f(x) : a) f(x) = x +x f(x+) f(x) = (x+) +(x+) (x +x) = x+ + = x++ f(x+) f(x) Thus lim = lim x++ = x+. b) f(x) = cos(ax), so that (at x = ) f() f() = cos(a). f(x+) f(x) = cos(a) = cos (a) (cos(a)+) = sin (a) (cos(a) + ) The relevant three limits are lim sin(a)/ = a, lim sin(a) = and lim /(cos(a)+) = /, so that lim sin (a) (cos(a)+) = lim sin(a) sin(a) cos(a)+ sin(a) = lim lim sin(a)lim cos(a)+ = ( a) (/) = ) Derivatives: a) f(x) = x (a+x), so that f (x) = (a+x) x(a+x) (a+x) = (a+x) x 6 (a+x) = a x 4 (a+x). 4 b) f(x) = sin(x) cos(x), so that f (x) = cos(x) cos(x) + sin (x) cos (x) = cos (x) = sec (x) or from table. c) f(x) = cos(ax), so that f (x) = asin(ax) from table. d) f(x) = cos (x)+sin (x) =, so that f (x) =. e) f(x) = cos (x) sin (x) = cos(x), so that f (x) = sin(x), or f (x) = sin(x)cos(x) sin(x)cos(x) = sin(x). f) f(x) = cos(x), so that f (x) = sin(x) from c). g)f(x) = a lnx = e lnalnx which is, by the way,x lna. f (x) = lna lna x elnalnx = x alnx. h) f(x) = x x = e xlnx, so that f (x) = (lnx)e xlnx + x x exlnx = (+lnx)x x. ) Stationary points etc. a) f(x) = x x, so that f (x) = x x x (x ) = x (x ) (x ). Sufficient for f (x) = : Assume x which is itself not a root, so x (x ) =, which has two solutions: Either x =, or x, so that x = and therefore x = /. Thus the stationary points are x = and x = /. Second derivative f (x) = 6x x 6x (x ) + x (x ) vanishes at x =, so that x = is neither a minimum or a maximum. It could be a point of inflection, to be tested below. At x = / the second derivative is 9/(/) (54/4)/(/4)+(54/8)/(/8) = 8 >, so thatx = / is a minimum. Test for point of inflection at x = : Find third derivative, f (x) = 6 x + terms with at least one factor x, thereforef () = 6 an = is a point of inflection (maximum [local] slope) and a stationary point.

2 Asymptotes: Vertical asymptotes (discontinuity) at x = : lim x + x x = + and lim x x x = There are no (simple) horizontal asymptotes. 5 minimum at x=/ point of inflection at x= 5 at x= b) f(x) = (x )/(x +4x 4), as a matter of habit: Roots of denominator (x+) =, sox = ±. f (x) = (x +4x 4) (x )(4x+4) (x +4x 4). Finwithf (x) = ;f well defined for x not one of the roots. Then(x )(4x+4) = (x +4x 4) is sufficient, so x 4x = which has two solutions x = or x = 4, so that x = after dividing by x. Thusx =, are the two stationary points. Second derivative f (x) = (4x+4) (x )4 (x +4x 4) + (x )(4x+4) (x +4x 4), which gives f () = /4 >, so that x = is a minimum. At x = one findsf () = /6 <, so that x = is a maximum. Asymptotes: Vertical asymptotes (discontinuity) atx = ± derived earlier. Horizontal asymptotes (x ± ) both are, because the denominator of f is of higher order in x than the numerator. To draw the graph, for x < the denominator is x +4x 4 = (x+) 6 > and numeratorx <, sof(x) < for allx <. For < x < denominator is negative, and so is numerator. For < x denominator is positive, and numerator flips sign from negative to positive at x=. at x=- / - minimum at x= at x= / horizontal asymptote at y= - maximum at x= -

3 4) Integrals a) xe 5x = 5 xe5x 5 e5x +C = 5 (x 5 )e5x +C b) e x sin(x) = e x sin(x) e x cos(x)+c = e x sin(x) e x cos(x) 9e x sin(x) + C Note that the last integral is identical to the original question. Reshuffle terms: e x sin(x) = ex (sin(x) cos(x))+c. c) π sin (x) = π sin (x)+cos (x) = π or π sin (x) = π ( cos(x)) = [ 4 sin(x)]π = π using cos(x) = sin (x). d) x = x du u = u+c = x +C using substution u = x and du = x e) I = cos(x) cos(x)+sin(x) = cos(x)(cos(x) sin(x)) (cos(x)+sin(x))(cos(x) sin(x)). Denominator is cos (x) sin (x) = cos(x), so I = cos(x) sin(x) = sin(x) + cos(x)+c. f) Note that x +x+ = (x+) = x+, so x+ x +x+ = x+ = [x +x] =. g) π/ sin 5 (x)cos(x) = u 5 du = /6 using u = sin(x) so du = cos(x). h) + x + x + 4x + 5x 4 + 6x 5 = [x + x + x + x 4 + x 5 +x 6 ] = 6 i) +x+x +4x +5x 4 +6x 5 = [x+x +x +x 4 +x 5 +x 6 ] = 6 j) +x+x +4x +5x 4 +6x 5 = [x+x +x +x 4 +x 5 +x 6 ] = or from h)-i) 5) Difficult Integrals a) +cos(x) = cos(x) cos (x) = sin (x) cos(x) sin (x) = cot(x)+ sin(x) +C by inspection OR t = tan(x/),tan(x) = t t,sin(x) = t,cos(x) = t, = dt +cos(x) = + t dt = dt = t+c = tan(x/)+c Note: cot(x)+ sin(x) = t t + t = t b) cos(x) = = cot(x) sin(x) +C = t t t +C = cot(x/)+c c) +sin(x) = sin(x) sin (x) = tan(x) cos(x) +C t = t = tan(x/). OR = + t dt = dt () = +C = an(x/) +C Note: tan(x) cos(x) = t t t = ( t) t = t, which seems to differ from the alternative result /()+C. It s all in the constant: + = t, so tan(x) = cos(x) d) sin(x) = tan(x)+ cos(x) +C +C = an(x/) + +C. = t t + t +C = t +C = t +C e) Use t = tan(x);sin (x) = t ;cos (x) = ; = dt +cos (x) = dt + = dt +t = +t dt

4 ( ( = tan t )+C = tan tan(x) )+C f) cos (x) = = cot(x)+c by inspection. sin (x) g) +sin (x) = dt + t = dt = / ( = tan ( tan(x))+c ) +t dt h) sin (x) = tan(x)+c by inspection tan (x) = /cos (X) i) sin (x) = ( x sin (x) ) noting that d sin (x) = x and f n f = f n df = n+ fn+ (x) j) ln(x) = (x) ln(x) = xln(x) x+c integration by parts. k) x n ln(x) = n+ (xn+ ) ln(x) = xn+ n+ ln(x) (n+) x n+ +C l) J(m,n) = π/ cos m (θ)sin n (θ)dθ J(,) = π/ dθ = π/ J(,) = π/ sin(θ)dθ = [ cos(θ)] π/ = J(,) = π/ sin (θ)dθ = π/ ( cos(θ))dθ = [θ/] π/ [ 4 sin(θ)]π/ = π/4 or J(,) = π/ sin (θ)+cos (θ)dθ = π/ because cos(θ) = sin(π/ θ) and so π/ cos (θ)dθ = π/ sin (π/ θ)dθ = π/ sin (θ)dθ = π/ sin (θ)dθ J(,) = π/ sin (θ)dθ = π/ ( cos (θ))sin(θ)dθ = +[(/)cos (θ)] π/ / J(,4) = π/ sin 4 (θ)dθ = 4 cos(θ), soj(,4) = 4 π/ 8 +cos(4θ)dθ = π [θ + 4 sin(4θ)]π/ = π J(4,) = π 6 π/ ( cos(θ)) dθ using cos (θ) = π/ cos(θ)+cos (θ)dθ = π 8 4 [sin(θ)]π/ + 6 using cos(θ) = sin(π/ θ), see J(,). J(,) = π/ cos(θ)dθ = [sin(θ)] π/ = J(,) = π/ cos(θ)sin(θ)dθ = [sin (θ)] π/ = using d dθ sin(θ) = cos(θ) and either substitute s = sin(θ) or consider f f = (f ). J(m,) = π/ cos m (θ)sin(θ) = m+ [cosm+ (θ)] π/ = m+ as above J(,n) = π/ sin m (θ)cos(θ) = n+ [sinn+ (θ)] π/ = n+ as above m)i = x (x +a ) = as integrand is odd. Note: Substitution y = x, so = x is valid only for x (, ] and x (,) separately (as the integrand is finite at x =, you can omit the single point x = ). This is obvious when considering x(y) = y, which covers only x [, ) and x(y) = y which covers only x (,]. x (x +a ) = x (x +a ) + x (x +a ) In the first integral on the right use x = y and = y integral use x = y and = y. I = y (y+a ) y + y (y+a ) y Note the limit of the second integral., in the second =

5 = (y+a ) + n) I = x x = y (y+a ) = (same integrand, swapped limits). ( y )y using y = x. I = y = ( ) y + ( ) ( y = ln( y ) ln( )+C = ln +C = ln ) x +C d Note: ln(x) = /x, d ln( x) = /x, so = ln()+c = ln( y)+c = ln( )+C all valid. Last one safest. o) ( ) = 6 ln x+ +C x 4 p) 4+x = tan(x/)+c see e) 8+x x = + x (x+)(x 4) = ( ) 6 x+ x 4