Help me double check my implicit differentiation problem?? Will give medal!

Question

anonymous · Answer

Please post your problem

anonymous · Answer

Thank you so much! Finding the slope of the tangent line at (4,0) for curve xy²+5tan(xy)-2y+x=4 @JuanitaM

anonymous · Answer

*curve xy^2 +5 tan (x y) - 2y + x = 4

anonymous · Answer

should it be (xy)^2 +5 tan (x y) - 2y + x = 4  ??

anonymous · Answer

or xy^2

anonymous · Answer

xy^2

anonymous · Answer

k

misty1212 · Answer

HI!!

misty1212 · Answer

your math teacher must not like you too much 
$$y^2+2xyy'+5\sec^2(xy)\left(y+xy'\right)=0$$ is a start

anonymous · Answer

Okay, if I did it correctly, I found it reduced to 5 sec ^ 2 (y (d y / d x ) ) + x = 4 ??

misty1212 · Answer

oh wait there was more!!

misty1212 · Answer

$$y^2+2xyy'+5\sec^2(xy)\left(y+xy'\right)-2y'+1=0$$

misty1212 · Answer

oh but $y=0$ whew, most goes away

misty1212 · Answer

$$5xy'-2y'+1=0$$

misty1212 · Answer

hmm i think you might have made an error with 
$$5\tan(xy)$$

anonymous · Answer

Okay, where is the y^2 coming from? I think I had expanded mine to :

1 * 2 y' + 5 sec ^ 2 ( xy ) - 2 + 1 ?
Could you show me what I did wrong?

misty1212 · Answer

sure, but first let me make sure i have it right

misty1212 · Answer

$$xy^2+5\tan(xy)-2y+x=4$$ right ?

anonymous · Answer

yes

misty1212 · Answer

ok so the derivative of $-2y$  is $-2y'$ and the derivative of $x$ is $1$ so we got the $-2y'+1$ out at the end yes?

anonymous · Answer

does y not act the same as the x variable where x^2 then becomes 2x^1?

anonymous · Answer

I am just not understanding why it wouldn't be -2(1)y^0.. which would just make it -2?

misty1212 · Answer

$y$ is to be considered as a function of $x$, just not explicitly written as one (hence "implicit diff)

misty1212 · Answer

think of it as 
$$xf^2(x)+5\tan(xf(x))-2f(x)+x=4$$

anonymous · Answer

Oh oh sorry I was just confusing myself. it is like replacing every y with a dy/dx or y'
sorry

misty1212 · Answer

so the first part requires the product rule 
the derivative of $$xf^2(x)$$ is $$f^2(x)+2xf(x)f'(x)$$

misty1212 · Answer

course it is a lot easier to write 
$$y^2+2xyy'$$

anonymous · Answer

That makes sense

anonymous · Answer

and -5 tan (xy) = -5 sec^2 (xy) ?

misty1212 · Answer

oh no

misty1212 · Answer

first off it is plus right? 
$$+5\tan(xy)$$

anonymous · Answer

yes, my mistake

misty1212 · Answer

then when you take the derivative, you do get 
$$5\sec^2(xy)$$ but the via the chain rule you have to multiply that by the derivative of $xy$

anonymous · Answer

so 5xy sec ^2 (xy) ?

misty1212 · Answer

which again requires the product rule 
the derivative of $xy$ is $y+xy'$

misty1212 · Answer

the derivative of $xy$ not $xy$

anonymous · Answer

well, (5y +5xy') sec^2 (xy)

anonymous · Answer

?

misty1212 · Answer

sure why not you can write it that way too

misty1212 · Answer

i wrote it like this $$y^2+2xyy'+5\sec^2(xy)\left(y+xy'\right)-2y'+1=0$$ which is a big mess until you replace y by zero and then almost everything goes away

anonymous · Answer

which simplifies to 5( xy' ) - 2y'+1 ?

misty1212 · Answer

yes

misty1212 · Answer

=0

anonymous · Answer

which becomes -1/18 once x is plugged in which equals about -.0556?

misty1212 · Answer

that i don't know, i didn't do it !

anonymous · Answer

Thank you so much! I was having a lot of trouble with the chain rule especially and this cleared it up a bit. Thanks!

misty1212 · Answer

yw $$\color\magenta\heartsuit$$