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implicit differentiation

Question: Find the slope of the curve xe^(x^2.y) + 3x^2 = 2y + 4 at the point (x, y) = (1, 0).

Is differentiating this to...

2x^2.e^(x^2.y)dy/dx + e^(x^2.y) + 6x = 2dy/dx

correct?

Used product rule on the first term.
Original post by Nightvision
Question: Find the slope of the curve xe^(x^2.y) + 3x^2 = 2y + 4 at the point (x, y) = (1, 0).

Is differentiating this to...

2x^2.e^(x^2.y)dy/dx + e^(x^2.y) + 6x = 2dy/dx

correct?

Used product rule on the first term.


No that's not correct, you also need the product rule on the exponent.
Original post by Nightvision
Question: Find the slope of the curve xe^(x^2.y) + 3x^2 = 2y + 4 at the point (x, y) = (1, 0).

Is differentiating this to...

2x^2.e^(x^2.y)dy/dx + e^(x^2.y) + 6x = 2dy/dx

correct?

Used product rule on the first term.


No, you went wrong when you did x(e^(x^2y))'
Reply 3
Avatar for Nightvision
Nightvision
OP
Original post by ghostwalker
No that's not correct, you also need the product rule on the exponent.


I tried that, do you know where I went wrong?

u=x v=e^(x^2.y)
du= 1 dv= 2xe^(x^2.y) dy/dx

u.dv + v.du
= 2x^2 e^(x^2.y)dy/dx + e^(x^2.y)
Original post by Nightvision
I tried that, do you know where I went wrong?

u=x v=e^(x^2.y)
du= 1 dv= 2xe^(x^2.y) dy/dx

u.dv + v.du
= 2x^2 e^(x^2.y)dy/dx + e^(x^2.y)


Your dv/dx is incorrect.

Try the chain rule on v and you'll see there is another product to evaluate.
Reply 5
Avatar for Nightvision
Nightvision
OP
Just tried the chain rule i'm getting the same thing :s-smilie:

v=e^(x^2.y)
let w = x^2.y

w=x^2.y
dw= 2x dy/dx
v= e^w
dv= e^w

dv/dx = 2xe^w dy/dx
replace the w with the power again and its the same thing

What am I doing wrong? :frown:
Original post by Nightvision
Just tried the chain rule i'm getting the same thing :s-smilie:

v=e^(x^2.y)
let w = x^2.y

w=x^2.y
dw= 2x dy/dx
v= e^w
dv= e^w

dv/dx = 2xe^w dy/dx
replace the w with the power again and its the same thing

What am I doing wrong? :frown:



w=x^2.y

So, dw/dx = 2xy + x^2 dy/dx
Reply 7
Avatar for Nightvision
Nightvision
OP
Original post by ghostwalker
w=x^2.y

So, dw/dx = 2xy + x^2 dy/dx


oh shizzle, product rule in a product role

doh

so dw/dx = 2xye^(x^2.y) + x^2.e^(x^2.y) dy/dx
(edited 13 years ago)
Reply 8
Avatar for Nightvision
Nightvision
OP
Final answer, gradient of slope = 7
Reply 9
Avatar for Focus08
Focus08
16
Original post by Nightvision
Question: Find the slope of the curve xe^(x^2.y) + 3x^2 = 2y + 4 at the point (x, y) = (1, 0).

Is differentiating this to...

2x^2.e^(x^2.y)dy/dx + e^(x^2.y) + 6x = 2dy/dx

correct?

Used product rule on the first term.


Math B problem set lol
Reply 10
Avatar for Nightvision
Nightvision
OP
Original post by Focus08
Math B problem set lol


Yup. An hour and 20 mins left to complete.

I suck at Math.
Original post by Nightvision
Final answer, gradient of slope = 7


Agreed.

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