3.)Find the derivative dy/dx:

i) 3y^3-4x^2y+xy=-5 ii)y^3+sinhxy^2=3/2

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

determine the equation of the tangent line x=1

Guest Feb 11, 2016

#1**0 **

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 :)

Melody Feb 11, 2016

#2**0 **

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.**

Guest Feb 11, 2016

#3**+5 **

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 :)

Melody Feb 11, 2016

#4**0 **

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 :)

Melody Feb 11, 2016

#9**+5 **

Best Answer

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

CPhill Feb 11, 2016

#10**0 **

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

It is fixed now though :)

Melody Feb 11, 2016

#12**+5 **

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 !!!

CPhill Feb 11, 2016