Derivate problem?

Question

Determine the derivative of the function. Then determine an equation of the line tangent to the graph of

f(x)=3x^2-5/(4x-5) at x=0

Please if you can show me the steps of your work so i understand how you got to the answer. thank you

Answer 1

f(x)=3x^2-5/(4x-5) and f'(x) = 6x -20/(4x-5)^2
f(0) = -5/(-5) = 1 and f'(0) = -20/25 = -4/5
The tangent has the slope -4/5 at (0 , 1)
The tangent is y = x + 1

Answer 2

first take the derivative w resp to x:

First term derivative is 6x [deriv of a*x^n is a*n*x^(n-1)]

Second term derivative is 20/(4x-5)^2

This is how we get that second term:

The deriv of a/f(x) is the same as derivative of a*f(x)^-1. Use the same formula as above and the chain rule df(u)/dx = df(u)/du * d(u)/dx where u is a function of x. In this case u=4x-5, and f(u) = -5/u.

So d(fu)/du = d(-5/u)/du =d(-5*u^-1)/du = 5*u^-2 = 5/u^2;
and du/dx = d(4x-5)/dx = 4

So df(u)/du*du/dx = 5/u^2 * 4, or 20/(4x-5)^2.

The full derivative is then 6x + 20/(4x-5)^2.

Set x=0 to evaluate the derivative at x=0; this value is the slope of the tangent line at x=0.

For the equation of the tangent:

The equation for a straight line is ax + b, where a is the slope and b is the value at x=0 (y intercept). The slope you get from the derivative computed above, the intercept from the value of the original function at x=0

You should be able to get the answer from this info; I won't do the whole solution for you. I hope I haven't made any errors and that you understand this.

After reading the first resonse, it is not clear what your equation is; if it really is the ratio of two polynomials as the first responder assumed, it should be written (3x^2-5)/(4x-5)

Answer 3

Well, start with your function:

(3x^2-5)/(4x-5)

To set up the quotient rule, set (3x^2-5) as u and (4x-5) as v. The quotient rule is (v'u-u'v)/(v^2).

u' is 6x through the power rule, v' is 4 by the same rule. Now plug and chug and simplify:

[4(3x^2-5)] - [6x(4x-5)]/(4x-5)^2

(12x^2 - 20 - 24x^2 + 30x) / (4x-5)^2

Finally simplifies to (12x^2 - 30x + 20) / (4x-5)^2

So that's your derivative. Now evaluate the original function at zero:

3(0)^2 - 5 / 4(0) - 5

You end up getting -5/-5 which equals 1.

Plug zero into the derivative function to get the slope at that point:

12(0)^2 - 30(0) + 20 / (4(0) - 5)^2

You end up getting 20/25, which equals .8, which, at that point, you can put a tangent line into point slope form:

.8 = (y - 1) / (x)
.8x = y - 1
.8x + 1 = y

And the problem is solved!

Answer 4

y = (3x^2-5)/(4x-5)
dy/dx = [(4x-5)(6x) - (3x^2-5)(4)]/[(4x-5)^2]
dy/dx = [(12x^2 - 30x +20)]/[16x^2 - 40x + 25)]

since dy/dx =is equal to the equation of the slope
the first derivative of an equation is the equation of the line tangent to the graph
to determine a slope at a particular point, just substitute the value of x & y from (x,y)

Answer 5

f(x) = (3x² - 5)/(4x - 5)

use the quotient rule: d/dx (u/v) = (v du/dx - u dv/dx)/v²
In this case, u = 3x² - 5 and v = 4x - 5
d/dx f(x) = [(4x - 5) d/dx (3x² - 5) - (3x² - 5) d/dx (4x - 5)]/(4x - 5)²

get derivatives
= [(4x -5)(6x) - (3x² - 5)(4)]/(4x - 5)²

distribute
= (24x² - 30x - 12x² + 20)/(4x - 5)²

simplify
= (12x² - 30x + 20)/(4x - 5)²

factor out
= 2(6x² - 15x + 10)/(4x - 5)²

The slope of tangent at x = 0 is
= 2(6(0)² - 15(0) + 10)/(4(0) - 5)²
= 2(0 - 0 + 10)/(0 - 5)²
= 2(10)/5²
= 20/25
= 4/5

^_^

Answer 6

First of all, please edit and make clear whether you mean
f(x) = 3x^2 - [ 5/(4x-5) ] or
f(x) = ( 3x^2 -5 ) / (4x-5)
You're already getting different interpretations of what you're trying to say.

Answer 7

What are you using this for? Are you in pre-cal or something?