Find the derivative of the following:
1=x^2+4xy+4y
My answer is 2x. Is this correct?
Also, solve: ln(x^2-5)=0
take e of both sides
x^2-5=1
x^2=6
x=sqrt6 Is this correct?
to find the derivative of 1=x^2+4xy+4y
I have to know if you are finding dy/dx or not
if so, then you have to use implicit differentiation
0 = 2x + 4x(dy/dx) + 4y + 4dy/dx
-2x - 4y = dy/dx(4x + 4)
dy/dx = (-2x-4y)/(4x+4)
= -(x+2y)/(2x+2)
for the last, you should have
x = ± √6
remember , we cannot take ln or log of a negative, but since the x is squared first, both ±√6 will work
To find the derivative of the given equation, we need to differentiate each term with respect to x.
The equation is 1 = x^2 + 4xy + 4y.
Differentiating x^2 with respect to x gives us 2x.
Differentiating 4xy with respect to x requires applying the product rule. The product rule states that for two functions u(x) and v(x), the derivative of their product is given by u'(x)v(x) + u(x)v'(x).
Let u(x) = 4x and v(x) = y.
Differentiating u(x) = 4x with respect to x gives us u'(x) = 4.
Differentiating v(x) = y with respect to x gives us v'(x) = dy/dx.
Applying the product rule, we have 4xy' + 4y.
Lastly, differentiating 4y with respect to x gives us 0, since y is independent of x.
Therefore, the derivative of the equation 1 = x^2 + 4xy + 4y is 2x + 4xy'.
So, your answer of 2x is not correct. The correct answer is 2x + 4xy'.