I need help with the following derivative probelms. They alll need the chain rule.

1) y = x/sqrt(x^4+2)

2) g(t) = 5 cos pi(t)

3) h(x) = ln(2x^3+3)

Thanks for any and all help.

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- Oct 26th 2007, 04:03 PMSMAlvarezDerivative problems
I need help with the following derivative probelms. They alll need the chain rule.

1) y = x/sqrt(x^4+2)

2) g(t) = 5 cos pi(t)

3) h(x) = ln(2x^3+3)

Thanks for any and all help. - Oct 26th 2007, 04:18 PMgalactus
Don't be put off by the chain rule. Sometimes it is presented in a complicated manner, but it's really the derivative of the outside times the derivative of the inside.

Here's the first one. Let's change it to a product and use the product rule.

It's a good idea to do that when possible. The product rule is easier to work with than the quotient rule in most cases.

$\displaystyle x(x^{4}+2)^{\frac{-1}{2}}$

$\displaystyle x\overbrace{(\frac{-1}{2})(x^{4}+2)^{\frac{-3}{2}}}^{\text{derivative of outside}}\underbrace{(4x^{3})}_{\text{derivative of inside}}+(x^{4}+2)^{\frac{-1}{2}}(1)$

Now, hammer it into shape with some algebra:

$\displaystyle \frac{-2x^{4}}{(x^{4}+2)^{\frac{3}{2}}}+\frac{1}{(x^{4}+2 )^{\frac{1}{2}}}$

$\displaystyle \frac{-2x^{4}}{(x^{4}+2)^{\frac{3}{2}}}+\frac{1}{(x^{4}+2 )^{\frac{1}{2}}}\cdot\frac{(x^{4}+2)}{(x^{4}+2)}$

$\displaystyle \boxed{\frac{2-x^{4}}{(x^{4}+2)^{\frac{3}{2}}}}$ - Oct 26th 2007, 04:21 PMqyzren
you can use quotient or product rule for 1st one.

you'll get -2x^4 /(2+x^4)^1.5 + 1/(2+x^4)^0.5 if you do it right.

2. just differentiate cos then differentiate the inside...

g'(t) = -5π*sin (πt)

3. if ln [f(x)], then d/dx ln [f(x)] = f'(x)/f(x)

h'(x) = 6x²/[2x³+3] - Oct 26th 2007, 04:30 PMSMAlvarez
Thanks guys, the last one was so easy!! Anyway I still dont understand the second. Thanks.

- Oct 26th 2007, 04:32 PMqyzren
if you have a function cos[f(x)], then d/dx cos[f(x)] = -f'(x)*sin[f(x)]

- Oct 26th 2007, 04:42 PMSMAlvarez
Im still confused on the second one. By the way pi = the symbol pi or 3.14

Thanks. - Oct 26th 2007, 04:47 PMqyzren
pi is just a constant.

d/dx cos[f(x)] = -f'(x)*sin[f(x)]

apply this to your question

d/dx 5cos(πt)

5 at the front is just a constant, so just leave it.

formula says (once you multiply both sides by 5) => 5*d/dx cos[f(x)] = -5*f'(x)*sin[f(x)] your f(x) in this case is πt, so f'(x) = π.

so sub it in...