determine the equation of the tangent line x=1

Not quite, Melody.......when you took the derivative of 3y^3,   it should be  9y^2 y'   instead of 6y^2 y'

ok I am fairly new at this but this is what I think is correct

i) 3y^3-4x^2y+xy=-5 

\(3y^3-4x^2y+xy=-5 \\ 9y^2\frac{dy}{dx}-4(2xy+x^2\frac{dy}{dx})+x\frac{dy}{dx}+y=0\\ 9y^2\frac{dy}{dx}-8xy-4x^2\frac{dy}{dx}+x\frac{dy}{dx}+y=0\\ 9y^2\frac{dy}{dx}-4x^2\frac{dy}{dx}+x\frac{dy}{dx}=8xy-y\\ (9y^2-4x^2+x)\frac{dy}{dx}=8xy-y\\ \frac{dy}{dx}=\frac{8xy-y}{ 9y^2-4x^2+x }\\\)

I'd like another mathematician to check this please :)

Derivative:3y^3-4x^(2y)+xy=-5

The derivative of y is y'(x):

(d/dx(x)) y+y'(x) 9 y^2+x y'(x)-8 x^(2 y) ((d/dx(log(x))) y+log(x) y'(x)) = d/dx(-5)

The derivative of x is 1:

1 y+x y'(x)+9 y^2 y'(x)-8 x^(2 y) ((d/dx(log(x))) y+log(x) y'(x)) = d/dx(-5)

The derivative of log(x) is 1/x:

y+x y'(x)+9 y^2 y'(x)-8 x^(2 y) (1/x y+log(x) y'(x)) = d/dx(-5)

The derivative of -5 is zero:

y+x y'(x)+9 y^2 y'(x)-8 x^(2 y) (y/x+log(x) y'(x)) = 0

Expand the left hand side:

y-8 x^(-1+2 y) y+x y'(x)-8 x^(2 y) log(x) y'(x)+9 y^2 y'(x) = 0

Subtract y-8 y x^(2 y-1) from both sides:

x y'(x)-8 x^(2 y) log(x) y'(x)+9 y^2 y'(x) = -y+8 x^(-1+2 y) y

Collect the left hand side in terms of y'(x):

(x-8 x^(2 y) log(x)+9 y^2) y'(x) = -y+8 x^(-1+2 y) y

Divide both sides by -8 x^(2 y) log(x)+x+9 y^2:

Answer: |y'(x) = (-y+8 x^(-1+2 y) y)/(x-8 x^(2 y) log(x)+9 y^2)

Derivative: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)

4: Sorry young person!. The computer software is unable to understand written questions!. If you can express it into something concrete to solve, then you might get somewhere.

4.) It is given that e^xy=x[(x+1)^3] / (x^2+1) ,where x>0

determine the equation  of the tangent line at  x=1

\(e^xy=\frac{x(x+1)^3}{(x^2+1)}\\ \mbox{Left hand side}\\ \frac{d}{dx}e^xy=e^xy+e^x\frac{dy}{dx}\\ \mbox{Right hand side}\\ Let\;\;u=x(x+1)^3\qquad \qquad \qquad \qquad and \qquad v=x^2+1\\ \qquad \;\;u'=(x+1)^3+3x (x+1)^2 \qquad and \qquad v'=2x\\ \frac{d}{dx}\frac{x(x+1)^3}{(x^2+1)}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[x(x+1)^3*2x]}{(x^2+1)^2}\\ \frac{d}{dx}\frac{x(x+1)^3}{(x^2+1)}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\so\\ e^xy+e^x\frac{dy}{dx}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\ \)

\(e^xy+e^x\frac{dy}{dx}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\ when\;\; x=1\\ ey+e\frac{dy}{dx}=\frac{2*[8+3*4]-[2*8]}{4}\\ ey+e\frac{dy}{dx}=\frac{40-16}{4}\\ ey+e\frac{dy}{dx}=6\\ \frac{dy}{dx}=\frac{6-ey}{e}\\\)

NOW

\( e^xy=\frac{x[(x+1)^3] }{ (x^2+1)}\\ When\;\;x=1\\ ey=\frac{8}{ 2}\\ y=\frac{4}{ e}\\ so\;\;when\;\;x=1\\ \frac{dy}{dx}=\frac{6-e*\frac{4}{e}}{e}\\ \frac{dy}{dx}=\frac{2}{e}\qquad \mbox{This is the gradient ofthe tangent at x=1}\\\)

So now for the equation of the tangent at (1, 4/e)

\(\frac{2}{e}=\frac{y-\frac{4}{e}}{x-1}\\ \frac{2(x-1)}{e}=\frac{ey-4}{e}\\ 2(x-1)=ey-4\\ 2x-2+4=ey\\ y=\frac{2x+2}{e}\\\)

I worked this out and coded it at the same time so it could be riddled with errors or it could just be plain wrong.   

OR maybe there is a much quicker way.....   idk ....    But that is my shot :)

Hi Guest

"4: Sorry young person!. The computer software is unable to understand written questions!. If you can express it into something concrete to solve, then you might get somewhere."

What computer software.

Aren't you doing these yourself??

I expect the asker has to do it by themselves without plugging it into any software ://

Mind you who needs software when you can just ask a mathematician to do it for you :)

3y^3-4x^2y+xy=-5 

9y^2y'   - 8xy - 4x^2y' + y  + xy '   =  0 

y ' [ 9y^2 - 4x^2 + x] =  8xy + y

y ' =  y [ 8x  - 1]  /  [ 9y^2 - 4x^2 + x ]

Edit :  I copied it incorrectly, the first time....!!!!!

Chris, I think you copied the question wrong.  It is 3y^3 not just y^3

Thanks, Melody.....I'll make the change....!!!

Great :)  Now our answers for question 1 are the same :)

Thanks Chris, I had already changed that but I must not have hit the final 'publish'

It is fixed now though :)

OK.....I see it, now.....

I'am going to assume that this is :

y^3+sinh(x)y^2=3/2         ......if so.....we have

3y^2y ' + cosh(x)y^2  + 2sinh(x)yy '  =  0

y ' [ 3y^2  + 2y sinh(x) ]   =   - cosh(x)y^2

y'  = -cosh(x)y^2 / ' [ 3y^2  + 2y sinh(x) ]  =  - y*cosh(x)  / [3y + 2sinh(x)]

With this level of math question......I'm surprised at your lack of parentheses/brackets !!!