# Thread: Derivatives, slope of tangent line.

1. ## Derivatives, slope of tangent line.

Find an equation of the tangent line to the the graph of the function f(x) =

(x^(1/2))(9-6x^2)

x

when x=4.

We note that, when x=4,
the corresponding point on the graph is given by (4,f(4))=(4, ___ ) and that the slope of the tangent line is f ′(4)=
Therefore the equation of the tangent line is y =

I'm not sure whether or not to cancel out the x's and have x^-1/2 at the top and then use the product rule. or use the quotient rule straightaway. Am I not seeing anything that's there?

I also get the slope 2.375

final equation I got (which is way off)

y = 2.375x-53

help would be appreciated!

2. See attachment

I'm not sure whether or not to cancel out the x's and have x^-1/2 at the top and then use the product rule. or use the quotient rule straightaway. Am I not seeing anything that's there?
As you would have to use the product rule anyway if you were to use the quotient rule I'd cancel and just use the product rule

3. Originally Posted by youmuggles
Find an equation of the tangent line to the the graph of the function f(x) =

(x^(1/2))(9-6x^2)

x

when x=4.

We note that, when x=4,
the corresponding point on the graph is given by (4,f(4))=(4, ___ ) and that the slope of the tangent line is f ′(4)=
Therefore the equation of the tangent line is y =

I'm not sure whether or not to cancel out the x's and have x^-1/2 at the top and then use the product rule. or use the quotient rule straightaway. Am I not seeing anything that's there?

I also get the slope 2.375

final equation I got (which is way off)

y = 2.375x-53

help would be appreciated!
You could distribute the $\displaystyle \sqrt{x}$ in the numerator?

$\displaystyle \frac{\sqrt{x} (9-6x^2)}{x} = \frac{9\sqrt{x} - 6x^{\frac{3}{2}}}{x}$

Using the quotient rule

$\displaystyle u = 9\sqrt{x} - 6x^{\frac{3}{2}} \: \rightarrow \: u' = \frac{9}{2\sqrt{x}} - 9\sqrt{x}$

$\displaystyle v = x \: \: \rightarrow \: \: v' = 1$

$\displaystyle f'(x) = \frac{x\left(\frac{9}{2\sqrt{x}} - 9\sqrt{x}\right) - (9\sqrt{x} - 6x^{\frac{3}{2}})}{x^2}$$\displaystyle = \frac{4.5\sqrt{x} - 9x^{\frac{3}{2}} - 9 \sqrt{x} + 6x^{\frac{3}{2}}}{x^2} = \frac{-4.5\sqrt{x} - 3x^{\frac{3}{2}}}{x^2}$

OR by simplifying

$\displaystyle 9x^{-\frac{1}{2}} - 6\sqrt{x}$

$\displaystyle f'(x) = -4.5x^{-\frac{3}{2}} - 3x^{-\frac{1}{2}}$

Note what happens if we multiply this expression by $\displaystyle \frac{x^2}{x^2}$

$\displaystyle \left(-4.5x^{\frac{3}{2}} - 3x^{-\frac{1}{2}}\right) \times \frac{x^2}{x^2} = \frac{-4.5x^{\frac{1}{2}} - 3x^{\frac{3}{2}}}{x^2}$

We see it's the same as above - in other words both methods are correct!

To find the slope find f'(4)

$\displaystyle f'(4) = -4.5\times 4 ^{-\frac{3}{2}} - 3 \times 4^{-\frac{1}{2}} = -\frac{4.5}{8} - \frac{3}{2} = -2.065$

4. the last term should be -6x^(5/2)