1. ## Need help please ! [urgent]

Hi there

1) If $\displaystyle y$ is a differentiable function of $\displaystyle x$ then the slope of the tangent to the curve $\displaystyle xy-2y+4y^2=6$ at the point where [tex]y=1[tex] is..?

2)Find the point on the graph of y=x^(1/2) between (1,1) and (9,3) at whtich the tangent to the graph has the same slop as the line through (1,1) and (9,3)

3) Let f and g be differential functions such that :
f(1) = 4
g(1) = 3
f'(3) = -5
f'(1)= -4
g'(1)= -3
g'(3) = 2

If h(x) = f(g(x)), then h'(1) =

4) Find the derivative and the equation of the tangent line to $\displaystyle 3x^2-3xy+2y^2=2$

5) Find the second derivative of $\displaystyle y^2=x^2-2x$

6) Use the tangent line approximation of y= square root of x at x=16 to find the approximate value of the square root of 17

And also, what does differentiable mean? How can you tell on a graph if a point is differentiable or not?

Thanks for the help

2. Originally Posted by NoiseandAttack
Hi there

1) If $\displaystyle y$ is a differentiable function of $\displaystyle x$ then the slope of the tangent to the curve $\displaystyle xy-2y+4y^2=6$ at the point where [tex]y=1[tex] is..?
first find the corresponding x-value when y = 1 from the given equation.

then differentiate implicitly and solve for y'. then plug in y = 1 and the corresponding x-value to solve

can you continue?

3. Originally Posted by Jhevon
first find the corresponding x-value when y = 1 from the given equation.

then differentiate implicitly and solve for y'. then plug in y = 1 and the corresponding x-value to solve

can you continue?
What I have a problem with is finding the derivative of the equation. I'm not really sure how. Can you explain please?

4. Originally Posted by NoiseandAttack
2)Find the point on the graph of y=x^(1/2) between (1,1) and (9,3) at whtich the tangent to the graph has the same slop as the line through (1,1) and (9,3)
find the slope of the line that passes through (1,1) and (9,3) (do you know how to do that?)

then, find the derivative of y = x^(1/2) and set it equal to the value you found above. solve for x, this will give you the x-value of the point you are after. then plug this x-value into the original function to solve for the corresponding y-value, and then you'll have your point

5. Originally Posted by NoiseandAttack
What I have a problem with is finding the derivative of the equation. I'm not really sure how. Can you explain please?
remember, when doing implicit differentiation, we keep track of what we are differentiating and what we are differentiating with respect to. here we are differentiating with repsect to x. so when we differentiate an x-term, we attach dx/dx to it (since we differentiated an x-term with respect to x). but derivative notations can function as fractions and this cancels and we get 1, so we don't write it. when we differentiate a y-term, we attach dy/dx (since we differentiated a y-term with respect to x). we justify doing this by the chain rule. you can look it up in your text for more info.

i will use y' to mean dy/dx to save on typing

$\displaystyle xy - 2y + 4y^2 = 6$

differentiating implicitly we get:

$\displaystyle y + x~y' - 2~y' + 8y~y' = 0$ ...........note that i used the chain rule for the first term

now, solve for y' and plug in the required x and y-values

6. Originally Posted by NoiseandAttack
3) Let f and g be differential functions such that :
f(1) = 4
g(1) = 3
f'(3) = -5
f'(1)= -4
g'(1)= -3
g'(3) = 2

If h(x) = f(g(x)), then h'(1) =
Let $\displaystyle h(x) = f(g(x))$

by the chain rule:

$\displaystyle h'(x) = f'(g(x)) \cdot g'(x)$

now, plug in x = 1 and solve

7. Originally Posted by NoiseandAttack
4) Find the derivative and the equation of the tangent line to $\displaystyle 3x^2-3xy+2y^2=2$
finding the derivative is very similar to the first question.

to find the tangent line, we need to know at what point. when you have the point, use the point-slope form:

$\displaystyle y - y_1 = m(x - x_1)$

where $\displaystyle m$ is the slope of the tangent line at the point $\displaystyle (x_1,y_1)$ and $\displaystyle (x_1,y_1)$ is a point the line passes through

just solve for y

8. Originally Posted by NoiseandAttack
5) Find the second derivative of $\displaystyle y^2=x^2-2x$
we can use implicit differentiation here, but i think it will be too much work

rewrite as $\displaystyle y = \pm \sqrt{x^2 - 2x}$

and differentiate twice using the chain rule. you will also need the product rule for the second derivative

9. Originally Posted by NoiseandAttack
6) Use the tangent line approximation of y= square root of x at x=16 to find the approximate value of the square root of 17
this is the formula for the linear approximation of a function.

$\displaystyle f(x) \approx f(a) + f'(a)(x - a)$

here $\displaystyle x$ is the value for which we want to find $\displaystyle f(x)$ for, but we can't because it does not have a nice answer, and $\displaystyle a$ is a point close to $\displaystyle x$ for which we do know the exact value of the function.

here $\displaystyle f(x) = \sqrt{x}$, $\displaystyle x = 17$ and $\displaystyle a = 16$

thus, $\displaystyle \sqrt{17} = f(17) \approx f(16) + f'(16)(17 - 16)$

now continue

10. Originally Posted by NoiseandAttack
And also, what does differentiable mean?
"differentiable" means we can differentiate it. that is, the function's derivative can be found, or exists at a certain point, or on some interval.

How can you tell on a graph if a point is differentiable or not?
recall how you defined the derivative. we started out with the slope formula for the equation of a line. thus, in layman's terms, a function is differentiable at a point, if it is approximately linear at that point. that is, if we zoom in close to a point, the function looks like a straight line. what this translates to is the following.

- we cannot find the derivative at points where a function takes a sharp turn, e.q. y = |x| at x = 0

- we cannot find the derivative of a function where it is discontinuous

11. Can you show the steps for finding the implicit derivative, please?

Thank you for all the other help, though.

12. Originally Posted by NoiseandAttack
Can you show the steps for finding the implicit derivative, please?

Thank you for all the other help, though.
what are you talking about? in post #5 i did show all the steps. all that was left for you to do is solve the equation for y'

13. Oohh, I'm so sorry.. I missed that !

Thanks for the help !

14. Originally Posted by NoiseandAttack
Oohh, I'm so sorry.. I missed that !

Thanks for the help !
ok. you're welcome

if you have any problems, say so