## YLynn Group Title how do I find dy/dx at: (x+2)^2-6(2y+3)^2=3, (1, -1) one year ago one year ago

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1. .Sam. Group Title

Looks like implicit differentiation, do you know how to approach this problem?

2. YLynn Group Title

a bit. yes, it is implicit differentiation

3. .Sam. Group Title

Here, most of the work involve chain rule only

4. YLynn Group Title

differentiate each term d/dx ( ln x ) + d/dx (ln (y)

5. YLynn Group Title

k

6. YLynn Group Title

how can you tell by looking at the equation

7. .Sam. Group Title

Differentiate (x+2)^2 you'll get 2(x+2) Differentiate -6(2y+3)^2 you'll get -12(2y+3)(2)(dy/dx) The 2 comes from chain rule, and dy/dx is differentiating with respect to x . Differentiate 3 is 0 Then you'll get $2(x+2)-12(2y+3)(2)\frac{dy}{dx}=0$ $2(x+2)-24(2y+3)\frac{dy}{dx}=0$ $\frac{dy}{dx}=\frac{x+2}{24 y+36}$

8. .Sam. Group Title

Because there's 2 terms together with a power, you'll have to use chain rule

9. .Sam. Group Title

Examples, 1) Differentiate 3(x+6)^2 dy/dx=6(x+6) 2) Differentiate 3(4x+2)^2 dy/dx=6(4x+2)(4) dy/dx=24(4x+2)

10. YLynn Group Title

ok. thanks for the examples. I understand those. For the question, the book gives an answer of: y'= 1/4 ?

11. Azteck Group Title

$\huge D_{x}[(x+2)^2-6(2y+3)^2]=0$ $\huge 2(x+2)-24(\frac{dy}{dx})(2x+3)=0$ Divide both sides by 2. $\huge (x+2)-12(\frac{dy}{dx})(2x+3)=0$ Make dy/dx the subject. $\huge \frac{dy}{dx}12(2x+3)=(x+2)$ $\huge \frac{dy}{dx}=\frac{x+2}{12(2x+3)}$ At (1, -1) $\huge \frac{dy}{dx}=\frac{1+2}{12[2(-1)+3]}$ $\huge \frac{dy}{dx}=\frac{3}{12(-2+3)}$ $\huge \frac{dy}{dx}=\frac{3}{12}$ $\huge =\frac{1}{4}$

12. Azteck Group Title

Use chain rule when differentiating in this situation.