Differentiation

**Macleef** · June 23rd 2008, 01:14 PM

Differentiate each of the following with respect to x. Simplify fully when indicated.

1) $y = e^{3x - 2} + ln(x^2 + 2x)$

Attempt:
$y' = 3e^{3x - 2} + ???$

I don't know how to differentiate for $ln(x^2 + 2x)$

Answer:
There is no answer in the book...

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2) $y = 3^{3x} - log_{10}(3x - 1)$

Attempt:
$y' = 3^{3x}ln3 - ???$

I don't know how to differentiate for $- log_{10}(3x - 1)$

Answer:
$y' = 3e^{3x - 2} + 3^{3x + 1}ln3$

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3) $y = \frac {e^{x}}{e^{x} - e^{-x}}$

Attempt:
$y' = \frac {e^{x}(e^{x} - e^{-x}) - (e^{x} - e^{-x}(-2))(e^{x})}{(e^{x} - e^{-x})^2}$
$y' = \frac {e^{x}(e^{x} - e^{-x}) - e^{x}(e^{x} + 2e^{-x})}{(e^{x} - e^{-x})^2}$

I'm not sure how to simplify it

Answer:
$y' = \frac{-2}{(e^{x} - e^{-x})^2}$

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4) $y = sin^{3}(3x - 1) - cos^2(x^2 - 1)$

Attempt:
I tried to use the product rule...

$y' = 3cos^2(3x - 1) + 3sin^{3} - 2sin(x^2 - 1) - 2cos^2$
$y' = 9cos^{2}x - 3cos^2x + 3sin^{3} - 2sin^2x + 2sin - 2cos^2x$

Answer:
$y' = 9sin^2(3x - 1)cos(3x - 1) + 4xcos(x^2 - 1)(sin(x^2 - 1)$

**Mathstud28** · June 23rd 2008, 01:20 PM

Originally Posted by Macleef

Differentiate each of the following with respect to x. Simplify fully when indicated.

1) $y = e^{3x - 2} + ln(x^2 + 2x)$

Attempt:
$y' = 3e^{3x - 2} + ???$

I don't know how to differentiate for $ln(x^2 + 2x)$

Answer:
There is no answer in the book...

----------

2) $y = 3^{3x} - log_{10}(3x - 1)$

Attempt:
$y' = 3^{3x}ln3 - ???$

I don't know how to differentiate for $- log_{10}(3x - 1)$

Answer:
$y' = 3e^{3x - 2} + 3^{3x + 1}ln3$

----------

3) $y = \frac {e^{x}}{e^{x} - e^{-x}}$

Attempt:
$y' = \frac {e^{x}(e^{x} - e^{-x}) - (e^{x} - e^{-x}(-2))(e^{x})}{(e^{x} - e^{-x})^2}$
$y' = \frac {e^{x}(e^{x} - e^{-x}) - e^{x}(e^{x} + 2e^{-x})}{(e^{x} - e^{-x})^2}$

I'm not sure how to simplify it

Answer:
$y' = \frac{-2}{(e^{x} - e^{-x})^2}$

----------

4) $y = sin^{3}(3x - 1) - cos^2(x^2 - 1)$

Attempt:
I tried to use the product rule...

$y' = 3cos^2(3x - 1) + 3sin^{3} - 2sin(x^2 - 1) - 2cos^2$
$y' = 9cos^{2}x - 3cos^2x + 3sin^{3} - 2sin^2x + 2sin - 2cos^2x$

Answer:
$y' = 9sin^2(3x - 1)cos(3x - 1) + 4xcos(x^2 - 1)(sin(x^2 - 1)$

$\frac{d}{dx}\bigg[\ln(u(x))\bigg]=\frac{u'(x)}{u(x)}$

So $\frac{d}{dx}\bigg[\log_a(u(x))\bigg]=\frac{1}{\ln(a)}\frac{d}{dx}\bigg[\ln(u(x))\bigg]=\frac{u'(x)}{\ln(a)u(x)}$

**Moo** · June 23rd 2008, 01:35 PM

Hello,

In general, the derivative of $\ln(ax+b)$ is $\frac{a}{ax+b}$ (this is chain rule)

1) For $\ln(x^2+2x)$ , note that this is $\ln[x(x+2)]=\ln x+\ln(x+2)$

Then, differentiate

2) $\log_a(x)=\frac{\ln(x)}{\ln(a)}$ , where $\ln(a)$ is obviously a constant

3) Just expand !
$e^xe^x=e^{2x}$ , using the rule for exponents.

4) Let $f(x)=\sin(3x-1)$ . $f'(x)=3 \cos(3x-1)$

You want to differentiate $(f(x))^3$ .

The chain rule would yield :

$3 \cdot f'(x) \cdot (f(x))^2$

etc... ^^

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