there are five question I do not have any idea to deal with.Any kind people can help?(just do what u know if you just know part of them) your little help will be a lot to me !)
Find the derivative dy/dx:
1a.)f(x)=(cosx)^(cosx) b.) 3y^3-4x^2y+xy=-5 c)y^3+sinhxy^2=3/2
2.)Let y=sin(n*sin^-1x), n>0.Show that \((1-x^2)y^"-xy^{'}+n^2y=0\)
3).It is give that \(e^{xy}=x(x+1)^{3}/(x^2+1), where, x>0. Determine, the ,equation,of,the,tangent,line,at,x=1\)
5.\(suppose that f is differentiable at x , show that \lim_{\Delta x\rightarrow 0}( f(x+\Delta x)-f(x-\Delta x))/(2*\Delta x)\)
Derivative:f(x)=(cosx)^(cosx)
Find the derivative of the following via implicit differentiation: d/dx(f(x)) = d/dx(cos^(cos(x))(x)) The derivative of f(x) is f'(x): f'(x) = d/dx(cos^(cos(x))(x)) Express cos^(cos(x))(x) as a power of e: cos^(cos(x))(x) = e^(log(cos^(cos(x))(x))) = e^(cos(x) log(cos(x))): f'(x) = d/dx(e^(cos(x) log(cos(x)))) Using the chain rule, d/dx(e^(cos(x) log(cos(x)))) = ( de^u)/( du) ( du)/( dx), where u = cos(x) log(cos(x)) and ( d)/( du)(e^u) = e^u: f'(x) = d/dx(cos(x) log(cos(x))) e^(cos(x) log(cos(x))) Express e^(cos(x) log(cos(x))) as a power of cos(x): e^(cos(x) log(cos(x))) = e^(log(cos^(cos(x))(x))) = cos^(cos(x))(x): f'(x) = cos(x)^(cos(x)) d/dx(cos(x) log(cos(x))) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = log(cos(x)): f'(x) = log(cos(x)) d/dx(cos(x))+cos(x) d/dx(log(cos(x))) cos^(cos(x))(x) The derivative of cos(x) is -sin(x): f'(x) = cos^(cos(x))(x) (cos(x) (d/dx(log(cos(x))))+-sin(x) log(cos(x))) Using the chain rule, d/dx(log(cos(x))) = ( dlog(u))/( du) ( du)/( dx), where u = cos(x) and ( d)/( du)(log(u)) = 1/u: f'(x) = cos^(cos(x))(x) (-(log(cos(x)) sin(x))+d/dx(cos(x)) sec(x) cos(x)) Simplify the expression: f'(x) = cos^(cos(x))(x) (d/dx(cos(x))-log(cos(x)) sin(x)) The derivative of cos(x) is -sin(x): f'(x) = cos^(cos(x))(x) (-(log(cos(x)) sin(x))+-sin(x)) Expand the left hand side: Answer: | | f'(x) = cos^(cos(x))(x) (-sin(x)-log(cos(x)) sin(x))
Derivative: 3y^3-4x^2y+xy=-5
Find the derivative of the following via implicit differentiation: d/dx(x y-4 x^2 y+3 y^3) = d/dx(-5) Differentiate the sum term by term and factor out constants: d/dx(x y)-4 d/dx(x^2 y)+3 d/dx(y^3) = d/dx(-5) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = y: -4 (d/dx(x^2 y))+3 (d/dx(y^3))+x d/dx(y)+d/dx(x) y = d/dx(-5) Simplify the expression: x (d/dx(y))-4 (d/dx(x^2 y))+3 (d/dx(y^3))+(d/dx(x)) y = d/dx(-5) The derivative of y is y'(x): -4 (d/dx(x^2 y))+3 (d/dx(y^3))+y'(x) x+(d/dx(x)) y = d/dx(-5) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^2 and v = y: 3 (d/dx(y^3))-4 x^2 d/dx(y)+d/dx(x^2) y+(d/dx(x)) y+x y'(x) = d/dx(-5) The derivative of y is y'(x): 3 (d/dx(y^3))+(d/dx(x)) y-4 (y'(x) x^2+(d/dx(x^2)) y)+x y'(x) = d/dx(-5) Using the chain rule, d/dx(y^3) = ( du^3)/( du) ( du)/( dx), where u = y and ( d)/( du)(u^3) = 3 u^2: 3 3 d/dx(y) y^2+(d/dx(x)) y+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) Simplify the expression: (d/dx(x)) y+9 (d/dx(y)) y^2+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) The derivative of y is y'(x): (d/dx(x)) y+y'(x) 9 y^2+x y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) The derivative of x is 1: 1 y+x y'(x)+9 y^2 y'(x)-4 ((d/dx(x^2)) y+x^2 y'(x)) = d/dx(-5) Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x: y+x y'(x)+9 y^2 y'(x)-4 (2 x y+x^2 y'(x)) = d/dx(-5) The derivative of -5 is zero: y+x y'(x)+9 y^2 y'(x)-4 (2 x y+x^2 y'(x)) = 0 Expand the left hand side: y-8 x y+x y'(x)-4 x^2 y'(x)+9 y^2 y'(x) = 0 Subtract y-8 x y from both sides: x y'(x)-4 x^2 y'(x)+9 y^2 y'(x) = -y+8 x y Collect the left hand side in terms of y'(x): (x-4 x^2+9 y^2) y'(x) = -y+8 x y Divide both sides by -4 x^2+x+9 y^2: Answer: | | y'(x) = (-y+8 x y)/(x-4 x^2+9 y^2)
y^3+sinhxy^2=3/2
Find the derivative of the following via implicit differentiation: d/dx(sinh^2(x y)+y^3) = d/dx(3/2) Differentiate the sum term by term: d/dx(sinh^2(x y))+d/dx(y^3) = d/dx(3/2) Using the chain rule, d/dx(sinh^2(x y)) = ( du^2)/( du) ( du)/( dx), where u = sinh(x y) and ( d)/( du)(u^2) = 2 u: d/dx(y^3)+2 d/dx(sinh(x y)) sinh(x y) = d/dx(3/2) Using the chain rule, d/dx(sinh(x y)) = ( dsinh(u))/( du) ( du)/( dx), where u = x y and ( d)/( du)(sinh(u)) = cosh(u): d/dx(y^3)+cosh(x y) d/dx(x y) 2 sinh(x y) = d/dx(3/2) Using the chain rule, d/dx(y^3) = ( du^3)/( du) ( du)/( dx), where u = y and ( d)/( du)(u^3) = 3 u^2: 2 cosh(x y) (d/dx(x y)) sinh(x y)+3 d/dx(y) y^2 = d/dx(3/2) The derivative of y is y'(x): 2 cosh(x y) (d/dx(x y)) sinh(x y)+y'(x) 3 y^2 = d/dx(3/2) Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = y: x d/dx(y)+d/dx(x) y 2 cosh(x y) sinh(x y)+3 y^2 y'(x) = d/dx(3/2) The derivative of y is y'(x): 2 cosh(x y) sinh(x y) (y'(x) x+(d/dx(x)) y)+3 y^2 y'(x) = d/dx(3/2) The derivative of x is 1: 3 y^2 y'(x)+2 cosh(x y) sinh(x y) (1 y+x y'(x)) = d/dx(3/2) The derivative of 3/2 is zero: 3 y^2 y'(x)+2 cosh(x y) sinh(x y) (y+x y'(x)) = 0 Expand the left hand side: 2 cosh(x y) sinh(x y) y+2 x cosh(x y) sinh(x y) y'(x)+3 y^2 y'(x) = 0 Subtract 2 y sinh(x y) cosh(x y) from both sides: 2 x cosh(x y) sinh(x y) y'(x)+3 y^2 y'(x) = -2 cosh(x y) sinh(x y) y Collect the left hand side in terms of y'(x): (2 x cosh(x y) sinh(x y)+3 y^2) y'(x) = -2 cosh(x y) sinh(x y) y Divide both sides by 2 x sinh(x y) cosh(x y)+3 y^2: Answer: | | y'(x) = -(2 cosh(x y) sinh(x y) y)/(2 x cosh(x y) sinh(x y)+3 y^2)
Let y=sin(n*sin^-1x), n>0.Show that: (1 - x^2)*y" - xy' + n^2y=0
n>0, y = sin(n sin^(-1)(x))
P.S. YOU HAVE TO RE-SUBMIT THE LAST TWO IN HAND-WRITING. DO NOT USE A MATH PACKAGE SUCH AS LA TEX.....ETC., BECAUSE WE CANNOT COPY AND PASTE THEM!.
