Find the derivative of the given function at the indicated point.

2 answers:

4 0

We're differentiating implicitly which means we're not going to find an explicit function y(x) before differentiating.

Simply power rule each term and don't forget to chain on the y. Since it's not x, we end up with this extra derivative thing y'.

$\rm 3x^2+2y^2=10$
$\rm 6x+4yy'=0$

Plug in x and y,

$\rm 6(1)+4\sqrt{3.5}~y'=0$

and solve for y',

$\rm y'(x)=\dfrac{-6}{4\sqrt{3.5}}$

Hope that helps.

Troyanec [42]2 years ago

3 0

3x^2 + 2y^2=10

d/d(x)= 6x+4y=0

Substitute x coordinate x=1

6(1)+4y'=0
6=-4y'
6/-4=y'
y'= -3/2

therefore the derivative of the function at x=1 is -1.5 or -3/2

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2 years ago

Write 3x^2-18x-6 in vertex form

Vedmedyk [2.9K]

The standard form of a quadratic equation is $\displaystyle{ y=ax^2+bx+c$ , while the vertex form is:

$y=a(x-h)^2+k$ , where (h, k) is the vertex of the parabola.

What we want is to write $\displaystyle{ y=3x^2-18x-6$ as $y=a(x-h)^2+k$

First, we note that all the three terms have a factor of 3, so we factorize it and write:

$\displaystyle{ y=3(x^2-6x-2)$ .

Second, we notice that $x^2-6x$ are the terms produced by $(x-3)^2=x^2-6x+9$ , without the 9. So we can write:

$x^2-6x=(x-3)^2-9$ , and substituting in $\displaystyle{ y=3(x^2-6x-2)$ we have:

$\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11]$ .

Finally, distributing 3 over the two terms in the brackets we have:

$y=3[x-3]^2-33$ .

Answer: $y=3(x-3)^2-33$