Exponential functions help

i have two questions i need help with if you could answer either that would be great,

1. Find the equation of the tangent to the curve y = ln(4x - 2) at the point (1, ln2)


2. Find the area enclosed by the curve y = e^x - x and the lines x= -1 and x = 1

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Originally Posted by perryman

i have two questions i need help with if you could answer either that would be great,

1. Find the equation of the tangent to the curve y = ln(4x - 2) at the point (1, ln2)

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$\displaystyle \frac{d}{dx}(ln[f(x)]) = \frac{f'(x)}{f(x)}$. Evaluate at the point 1.
You can then use this as $\displaystyle m$ in the equation of a straight line

Spoiler:

$\displaystyle \frac{d}{dx}(ln[f(x)]) = \frac{2}{2x-1} $

At point $\displaystyle f'(1) = 2$. $\displaystyle y-ln(2) = 2(x-1)$

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2. Find the area enclosed by the curve y = e^x - x and the lines x= -1 and x = 1

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$\displaystyle \int^1_{-1}(e^x-x)\,dx $

Spoiler:

$\displaystyle e - \frac{1}{e}$

alright this shouldn't be too difficult then. The definition of a derivative is the slope of the tangent line at any point in the graph. So, by finding the derivative of the function (evalutated at your point) should give you the slope of your tangent line at that point (x=1).

Then by using the equation for a linear equation

y=mx+b, plug in your slope (m), and your values of x and y given from your point, to solve for b.

thanks for the help everyone, if you guys could help me with one final thing that would be great......

find the coordinates of any stationary points on the curve y = x - e^(x-1)

i presume you need to use the product rule?

Quote:

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Originally Posted by perryman

thanks for the help everyone, if you guys could help me with one final thing that would be great......

find the coordinates of any stationary points on the curve y = x - e^(x-1)

i presume you need to use the product rule?

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I don't see that y is a product of functions .... y is a difference of two functions so differentiate each function in the usual way.


Edit: You already asked this question here: http://www.mathhelpforum.com/math-he...ions-help.html. Thread closed.