• Jul 10th 2006, 03:59 AM
Identify any general rules of calculus that you use

i) f(x) = tan(3x)In(sin(3x))

ii) f(x) = e^-sqrt(x) divided by x^3 +1
• Jul 10th 2006, 05:26 AM
galactus
For the first, use the product rule:

$\displaystyle tan(3x)ln(sin(3x))$

First times derivative of second plus second times derivative of first.

$\displaystyle tan(3x)\frac{d}{dx}[ln(sin(3x))]+ln(sin(3x))\frac{d}{dx}[tan(3x)]$

$\displaystyle tan(3x)(3cot(3x))+ln(sin(3x))(3sec^{2}(3x))$

I'll let you tidy it up a bit.
• Jul 10th 2006, 05:54 AM
malaygoel
Here is second.
Your function is $\displaystyle \frac{e^{-\sqrt{x}}}{x^3 + 1}$
Its differentiation is(using quotient rule)
$\displaystyle \frac{3x^2(e^{-\sqrt{x}})-(e^{-\sqrt{x}})(\frac{-1}{2\sqrt{x}})}{(x^3 + 1)^2}$
It may be useful to you:
If u,v,w,... are functions of x then
$\displaystyle \frac{d(uvw...)}{uvw...}=\frac{du}{u} + \frac{dv}{v} + \frac{dw}{w} + ...$

Keep Smiling
Malay
• Jul 10th 2006, 06:19 AM
Soroban

Quote:

Identify any general rules of calculus that you use

$\displaystyle 1)\;\;f(x) \,= \,\tan3x\cdot\ln(\sin3x)$

We have: .$\displaystyle f'(x) \;= \;\tan3x\cdot[\ln(\sin3x)]' + [\tan3x]'\cdot\ln(\sin3x)$ . . . product rule

. . . . . . . .$\displaystyle f'(x) \;= \;\tan3x\cdot\left[\frac{\cos3x}{\sin3x}\cdot3\right] \,+ \,\left[3\sec^23x\right]\cdot\ln(\sin 3x)$ . . . chain rule

Simplify: .$\displaystyle f'(x)\;=\;\underbrace{\tan 3x\cdot\cot3x}\cdot3 \,+ \,3\sec^23x\cdot\ln(\sin 3x)$
. . . . . . . . . . . . . . reciprocals

Therefore: .$\displaystyle f'(x)\;=\;3 + 3\sec^23x\ln(\sin3x)$

Quote:

$\displaystyle 2)\;\;f(x) \,= \,\frac{e^{-\sqrt{x}}}{x^3+1}$

This one is truly messy!

We have: .$\displaystyle f(x)\;=\;\frac{e^{-x^{\frac{1}{2}}}}{x^3 + 1}$

Then: .$\displaystyle f'(x) \;= \;\frac{(x^3+1)\left[e^{-x^{\frac{1}{2}}}\right]' - \left(e^{-x^{\frac{1}{2}}}\right)\left[x^3+1]'}{(x^3+1)^2}$ . . . quotient rule

$\displaystyle f'(x) \;= \;\frac{(x^3+1)e^{-x^{\frac{1}{2}}}\left(-\frac{1}{2}x^{-\frac{1}{2}}\right) \,- \,\left(e^{-x^{\frac{1}{2}}}\right)(3x^2)}{(x^3+1)^2}$ . . . chain rule

Now we simplify . . . (do we hafta?)

We have: .$\displaystyle f'(x) \;= \; \frac{-\frac{x^3+1}{2e^{x^{\frac{1}{2}}}x^{\frac{1}{2}}} - \frac{3x^2}{e^{x^{\frac{1}{2}}}}}{(x^3 + 1)^2}$

Multiply top and bottom by $\displaystyle 2e^{x^{\frac{1}{2}}}x^{\frac{1}{2}}:\;\;f'(x) \;= \;\frac{-(x^3+1) - 3x^2\cdot2x^{\frac{1}{2}}}{2e^{x^{\frac{1}{2}}}x^{ \frac{1}{2}}(x^3 + 1)^2}$

Therefore: .$\displaystyle f'(x) \;= \;-\frac{x^3 \,+ 1 + 6x^2\sqrt{x}}{2e^{\sqrt{x}}\sqrt{x}(x^3+1)^2}$

[Disclaimer: not responsible for egregious errors.]

• Jul 11th 2006, 01:57 PM