Derivatives

**Amanda-UK** · August 15th 2007, 04:29 AM

Hello would gratefully appreciate help on the following question, also would you be able to guid me through how you got the answers
thank you
Amanda x

Find dy/dx of the following:

1) y=3^(3x-1) + (1-2x)In(1-2x)

2) y= 2x-1 -log4(x+1)
3x+4

3)
y= e^x (2x+1)^1/2

4) intergrate and differentiate:
xIn(x+1)

**topsquark** · August 15th 2007, 05:34 AM

Originally Posted by Amanda-UK

Find dy/dx of the following:

1) y=3^(3x-1) + (1-2x)In(1-2x)

I presume that "In(1 - 2x)" is "ln(1 - 2x)"?

Note that $\frac{d}{dx}ln(x) = \frac{1}{x}$ .
Also $\frac{d}{dx}a^x = ln(a) \cdot a^x$ .

Thus, using the chain rule:
$y = 3^{3x - 1} + (1 - 2x)ln(1 - 2x)$

$\frac{dy}{dx} = \left ( ln(3x - 1) \cdot 3^{3x - 1} \right ) \cdot 3 + (-2) \cdot ln(1 - 2x) + (1 - 2x) \cdot \left ( \frac{1}{1 - 2x} \right ) \cdot (-2)$

$\frac{dy}{dx} = 3 \cdot ln(3x - 1) \cdot 3^{3x - 1} - 2 \cdot ln(1 - 2x) -2$

-Dan

**topsquark** · August 15th 2007, 05:43 AM

Originally Posted by Amanda-UK

2) y= 2x-1 -log4(x+1)
3x+4

I presume:
$y = \frac{2x - 1}{3x + 4} - log_4(x + 1)$

I, personally, never remember the formula for taking the derivative of a log function. What I do remember is the change of base formula:
$log_b(x) = \frac{log_a(x)}{log_a(b)}$
(This comes in very handy doing logs on my calculator.)

So I'm going to convert $log_4(x)$ into a $ln(x)$ term, which we know how to take the derivative of.

$log_4(x) = \frac{ln(x)}{ln(4)}$

Thus
$\frac{d}{dx}log_4(x) = \frac{d}{dx} \left ( \frac{ln(x)}{ln(4)} \right ) = \frac{1}{ln(4)} \cdot \frac{1}{x}$

So back to your problem:
$y = \frac{2x - 1}{3x + 4} - log_4(x + 1)$

$\frac{dy}{dx} = \frac{(2)(3x + 4) - (2x - 1)(3)}{(3x + 4)^2} - \frac{1}{ln(4)} \cdot \frac{1}{x + 1} \cdot 1$

$\frac{dy}{dx} = \frac{11}{(3x + 4)^2} - \frac{1}{ln(4)} \cdot \frac{1}{x + 1}$

-Dan

**topsquark** · August 15th 2007, 05:46 AM

Originally Posted by Amanda-UK

Find dy/dx of the following:
3)
y= e^x (2x+1)^1/2

Note that $\frac{d}{dx}e^x = e^x$ .

Thus by the product rule:
$\frac{dy}{dx} = e^x \cdot (2x + 1)^{1/2} + e^x \cdot \frac{1}{2} \cdot (2x + 1)^{-1/2} \cdot (2)$

$\frac{dy}{dx} = e^x \cdot (2x + 1)^{1/2} + e^x \cdot \frac{1}{(2x + 1)^{1/2}}$

-Dan

**topsquark** · August 15th 2007, 06:02 AM

Originally Posted by Amanda-UK

Find dy/dx of the following:
4) intergrate and differentiate:
xIn(x+1)

First the derivative:
$\frac{d}{dx} (x \cdot ln(x + 1) ) = (1) \cdot ln(x + 1) + x \cdot \frac{1}{x} = ln(x + 1) + 1$

Now the integral. I'm going to do this using integration by parts:
$\int u dv = uv - \int v du$

So
$\int x ln(x + 1) dx$ :
I'm going to set
$u = ln(x + 1) \implies du = \frac{dx}{x + 1}$
$dv = x dx \implies v = \frac{1}{2}x^2$

Thus
$\int x ln(x + 1) dx = \frac{1}{2}x^2 \cdot ln(x + 1) - \int \left ( \frac{1}{2}x^2 \right ) \left ( \frac{dx}{x + 1} \right )$

Which leaves us with the $\int \frac{x^2}{x + 1} dx$ , which you can do with partial fractions, or any other method of your choosing.

-Dan

**Amanda-UK** · August 15th 2007, 06:27 AM

thank you as helped me a great deal

**Krizalid** · August 15th 2007, 08:42 AM

Originally Posted by topsquark

$\int \frac{x^2}{x + 1} dx$

This is the simplest method

$\int\frac{x^2}{x+1}\,dx=\int\frac{x^2-1}{x+1}\,dx+\int\frac1{x+1}\,dx=\int(x-1)\,dx+\int\frac1{x+1}\,dx$

Regards.

K.

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