Expert: Alon Mandes - 10/5/2008

Question
1. Suppose variables x and y  are related by:

xy+2sin(y^2+1)=sin(x^2y^2)

Find an expression in terms of both x and y for the derivative y'(x).

Ok so i guess we use implicit differentiation? but, since we cant make y the subject, we differentiate each side of the equation to get,

y'(x)= y = 2xy^2coz(x^2y^2)

Now im lost! I dont understand the principle behind implicit differentiation, although i have notes on it...can you provide a simple explanation?
thankyou

Also, to differentiate:
h(t)= sin(2t^2-6)

i used the product rule-uv'+vu'

h'(t)= sin4t + cos(2t^2-6)

is this correct?

thanks again :)

Answer
Sarah, what you calculated concerning h'(t) is not correct at all :
If h(t)=sin(2t^2-6) then h'(t)=cos(2t^2-6)*(2t^2-6)'=
cos(2t^2-6)*(4t)=4tcos(2t^2-6).
You see that the rule is sin(f(t))'=cos(f(t))*f'(t).
Now , about implicit differentiation :
1st, implicit functions are function that x & y are related in such
a fashion that we can't get y as function of x, say y=f(x). Due to that, our traditional method of deriving is co longer valid. Instead
there is another proper method, which is called implicit differentiation. This method implies that x & y both should be derived with respect to x. Hence, x'=1 & y'=y'. For example :
x^2+y^2=1, the implicit differentiation will be : 2x+2yy'=0 -->
y'=-x/y. Another complicated example :
If the implicit function is Ln(x+y)=cos(y)+y^2+xy then the implicit
differentiation will be : (1+y')/Ln(x+y)=y'*sin(y)+2yy'+(xy)'
Note that the last expression will be treated as a product rule-uv'+vu'. Thus we well get :
(1+y')/Ln(x+y)=y'*sin(y)+2yy'+x'y+xy' ("x'=1")
(1+y')/Ln(x+y)=y'*sin(y)+2yy'+y+xy'
(1+y')/Ln(x+y)=y'(sin(y)+2y+x)+y
& from here , you can express y' as function of x & y.

Now let's get back to our exercise :
xy+2sin(y^2+1)=sin(x^2y^2)
Let's differentiate :
y+xy'+2cos(y^2+1)*2yy'=cos(x^2y^2)*(x^2y^2)' (" let's use product rule")
y+xy'+2cos(y^2+1)*2yy'=cos(x^2y^2)*(2xy^2+2yy'x^2)
y+xy'+4yy'cos(y^2+1)=2xy^2*cos(x^2y^2)+2yy'x^2*cos(x^2y^2)
y+xy'+4yy'cos(y^2+1)-2xy^2*cos(x^2y^2)-2yy'x^2*cos(x^2y^2)=0
y'[x+4ycos(y^2+1)-2yx^2*cos(x^2y^2)]+y-2xy^2*cos(x^2y^2)=0
y'=[2xy^2*cos(x^y^2)-y] / [x+4ycos(y^2+1)]

Any question, I'm here.
Alon.