Thread: Calculus boggles my brains :[

1. Calculus boggles my brains :[

Hey, can you guys explain how to do these problems? I'm stumped!

1) If f(x)=sinx and g(x)= (x+2)^(1/2) then the derivative of y=g(f(x)) at x=pi is ...

2) The lim as h -> 0 ( tan 4(x+h) - tan(4x) )/ h is....

3) Find the derivative of y=(3x^4)e^3x

4) If ln (4x) = e^y, what is dy/dx in terms of x?

5) If y = ln x, then the third derivative is...

7) If y = (g(x))^3 find y'(2) when g(2)=2 and g'(2)=10

8) Consider the curve defined by xy^2-x^3y=6
Find all points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

2. Originally Posted by NoiseandAttack
1) If f(x)=sinx and g(x)= (x+2)^(1/2) then the derivative of y=g(f(x)) at x=pi is ...
If $y=g[f(x)]$

differentiate
$y\prime=\frac{d}{dx}g[f(x)]$

So differentiate g(x) and apply the chain rule
$y\prime=g\prime[f(x)]f\prime(x)$

So just find g'(x) and f'(x), then plug f(x) into g'(x) and multiply that by f'(x)

3. Originally Posted by angel.white
If $y=g[f(x)]$

differentiate
$y\prime=\frac{d}{dx}g[f(x)]$

So differentiate g(x) and apply the chain rule
$y\prime=g\prime[f(x)]\prime(x)$

So just find g'(x) and f'(x), then plug f(x) into g'(x) and multiply that by f'(x)
I still don't seem to be getting the right answer

my work:
g'(x) = 1/2(x+2)^-.5
= (x+2)/ sqrt of 2
f'(x) = cosx

(cos x) + 2 / sqrt of 2 * cos x

cos pi +2 / sqrt of 2 * cos pi
-1/ sqrt of 2

4. error

NoiseandAttack,

$g'(x)=\tfrac{1}{2}(x+2)^{-\tfrac{1}{2}}$
which equals $\frac{1}{2 \sqrt{x+2}}$, not $\frac{x+2}{\sqrt{2}}$.

--Kevin C.

5. Error

Further, it's $y'=g'(f(x))f'(x)$, so you should be putting $f(x)=\sin x$ into g', not $f'(x)=\cos x$.

--Kevin C.

6. $f(x)=(sin\:x +2)^{\frac{1}{2}}$
$f'(x)=\frac{1}{2}(sin\:x+2)^{-\frac{1}{2}}(cos\:x)$
$f'(\pi)=\frac{1}{2}(0+2)^{-\frac{1}{2}}(-1)$
$f'(\pi)=-\frac{1}{2\sqrt{2}}$

$y=(3x^4)e^{3x}$
$y'=(12x^3)(e^{3x})+(3e^{3x})(3x^4)$
$y'=3e^{3x}x^3(3x+4)$

$ln(4x)=e^y$ to get this in terms of x, you have to ln both side
So... $y=ln(ln(4x))$
$ln(4x)=e^y$ ...use implict to find y'
$\frac{4}{4x}=e^y\frac{dy}{dx}$
$\dfrac{dy}{dx}=\frac{1}{xe^y} = \frac{1}{xe^{ln(ln(4x))}}$

$y=ln\:x$ log differentiation
$y'=\frac{1}{x}$ quotient rule
$y''=-\frac{1}{x^2}$ quotient rule
$y'''=\frac{2}{x^3}$

$f(x)=(g(x))^3$
$f'(x)=3(g(x))^2\:g'(x)$
$f'(2)=3(g(2))^2\:g'(2)$
$f'(2)=3(2)^2(10)$
$f'(2)=120$