find the slope of the tangent to the curve xy^2 +x^2y=2 at the point (1,1) ?

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find the slope of the tangent to the curve xy^2 +x^2y=2 at the point (1,1) ?

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Answers (5)

asad.taj

The given equation is
xy²+x²y = 2
differentiate it w.r.t x
x.2ydy/dx +y² + 2x.y+ x².dy/dx =0
(2xy + x² )dy/dx = -2xy - y²
dy/dx = (-2xy -y²)/(2xy+x²)

therefore dy/dx at (1, 1) = (-2.1.1 -1²)/(2.1.1+1²)
= -3/3 = -1

Posted 226 day ago

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davidmcy

In case the original post actually intended:

x(y^2) + x^(2y) = 2

we can handle that one too, like this:

Differentiating implicitly and with a logarithmic derivative gives:
x(y^2)' + x'(y^2) + 2(x^(2y))((y/x) + y'ln(x)) = 0

Taking the indicated derivatives and simplifying:
2xyy' + (y^2) + 2(x^(2y))((y/x) + y'ln(x)) = 0

Solving for y' with algebra:
y' = - ((y^2) + 2(x^(2y))(y/x)) / (2xy + 2(x^(2y))ln(x)))

So, the required slope, y'(1,1), is found by substituting x=1 and y=1 into this last formula, giving:
y'(1,1) = -(1+2)/(2+0) = -3/2

The point-slope form of the equation of the tangent line would be:
(y-1) = (-3/2)(x-1)

Posted 227 day ago

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pecks_tutoring

I'll assume the equation is:

xy² + x²y = 2

Use implicit differentiation to get dy/dx. Once you have dy/dx, you can plug any point into dy/dx to get the slope of the line tangent to the curve at that point.

The derivative is:
x*2y*dy/dx + 1*y² + 2xy + x²dy/dx = 0

Which is the same as:
2xy*dy/dx + x²*dy/dx + y² + 2xy = 0

To solve for dy/dx:
(2xy+x²)*dy/dx +y² + 2xy = 0

Subtract y² + 2xy from both sides to get
(2xy+x²)*dy/dx = - y² - 2xy

Divide both sides by 2xy+x² to get:
dy/dx = (- y² - 2xy)/(2xy+x²)

Now plug in x = 1, y = 1 to find the slope of the line tangent to the curve at (1,1)
dy/dx = slope = (- (1²) - 2*1*1)/(2*1*1+1²) = (-1-2)/(2+1) = -3/3 = -1

Math, statistics, computer science and business tutoring at peckstutoring.com

Posted 227 day ago

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math_fanatic

Student,
is the equation x(y^2) + (x^2)y = 2 or x(y^2) + x^(2y) = 2?
Please clarify.
Thanks!

Posted 227 day ago

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Catherine

Definition: equation of the tangent to the curve y(x) at the point (x0,y0)
f(x)=y'(x0,y0)*x+C
Here C- constant, y'(x0,y0)- derivative(slope) y(x) at the point (x0,y0)

a) y=1/sqrt x, y'=(-1/2)x^(-3/2)=(-1/2)a^(-3/2)

b) y'(x)= (-1/2)x^(-3/2)= -1/16.
f(x)=(-1/16)x+C
(4,1/2): 1/2= - 1/4+C
C=3/4

f(x)=(-1/16)x+3/4

Answer: a) slope is (-1/2)a^(-3/2)
b) the tangent f(x)=(-1/16)x+3/4

does this help....from Yahoo answers ?

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