# Calculus Questions

• October 9th 2008, 09:39 AM
midsummer
Calculus Questions
Hi I need help with these questions:

1.Write the composite function in the form f(g(x)). then find the derivative dy/dx. y=sin square root x.

2. Find the first and second derivatives of the function y=4x/ square root (X+1)

3. find dy/dx by implicit differentiation
square root (x+y)=1 +x^2 y^2
• October 9th 2008, 09:50 AM
watchmath
1) $f(x)=sin x, g(x)=\sqrt{x}$
by chain rule: $dy/dx=f'(g(x))g'(x)$. Try by yourself first!
• October 9th 2008, 09:56 AM
midsummer
I got cos square root x/ 2 square root x. Not sure if this is correct. :/
• October 9th 2008, 10:12 AM
watchmath
Quote:

Originally Posted by midsummer
I got cos square root x/ 2 square root x. Not sure if this is correct. :/

You need to be confidence on yourself (Clapping)(Clapping) (Hi)

For number 2, use quotient rule.

For number 3, you have
$\sqrt{x+y}=1+x^2y^2$

I'll do the LHS and you do the RHS
$
\frac{d}{dX}(x+y)^{1/2}=\frac{1}{2} (x+y)^{-1/2}\cdot \frac{d}{dx} (x+y)=\frac{1}{2} (x+y)^{-1/2} (1+\frac{dy}{dx})$

After you do the RHS, then solve the equation for dy/dx
• October 9th 2008, 10:37 AM
midsummer
So is my answer to the first question correct?

As for number 2. I got y= 2x+1/ 2(x+1)^1/2. Not sure about that :/

As for number 3. I am still lost :/

It's not that I lack confidance, I am trying to get better with practice and it's hard when I get stuck on questions. I really appreciate the help (Happy)(Hi)
• October 9th 2008, 10:43 AM
Chris L T521
Quote:

Originally Posted by midsummer
3. find dy/dx by implicit differentiation
square root (x+y)=1 +x^2 y^2

In implicit differentiation, you treat x and y as functions of x.

So when you differentiate $\sqrt{x+y}=1+x^2y^2$, you need to apply chain rule to $\sqrt{x+y}$ and product rule to $x^2y^2$

I'll show the initial first step, and then I'll let you simplify this.

$\frac{\,d}{\,dx}\left[\sqrt{x+y}\right]=\frac{\,d}{\,dx}\left[1+x^2y^2\right]\implies \tfrac{1}{2}(x+y)^{-\frac{1}{2}}\cdot\left(1+\frac{\,dy}{\,dx}\right)= 2xy^2+2x^2y\frac{\,dy}{\,dx}\implies\dots$

From here, simplify a little bit, but the important part is to group the terms that contain $\frac{\,dy}{\,dx}$ together. Then we can easily solve for $\frac{\,dy}{\,dx}$.

Can you take it from here?

--Chris