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