What is the second derivative of f(x) = x√(5-x)?

[frosty8688]

1. Find the second derivative of the following function
2. $f(x) = x\sqrt{5-x}, f'(x) = \frac{10-3x}{2\sqrt{5-x}}$
3. $f"(x)=\frac{-3*2\sqrt{5-x}-(10-3x)(2\sqrt{5-x})*-x}{(2\sqrt{5-x})^{2}}$ Here is where I get lost.

 

[stunner5000pt]

$f"(x)=\frac{-3*2\sqrt{5-x}-(10-3x)(2\sqrt{5-x})*-x}{(2\sqrt{5-x})^{2}}$ Here is where I get lost. [/b][/QUOTE]

There is a slight error in this step. When you differentiated the denominator in the second term, you have not subtracted 1 from the 1/2 power. This is what I got:

$$\frac{-3(2\sqrt{5-x})-(10-3x)(2)\frac{1}{2}(5-x)^{-1/2}(-1)}{(2\sqrt{5-x})^2}$$

 

[Mark44]

1. Find the second derivative of the following function



2. $f(x) = x\sqrt{5-x}, f'(x) = \frac{10-3x}{2\sqrt{5-x}}$



3. $f"(x)=\frac{-3*2\sqrt{5-x}-(10-3x)(2\sqrt{5-x})*-x}{(2\sqrt{5-x})^{2}}$ Here is where I get lost.

Also, it's probably simpler to write f'(x) as a product rather than a quotient, and write the radical in exponent form. That way you can use the product rule to get f''(x). I almost always prefer to use the product rule over the quotient rule, because the latter is a bit more complicated, making it easier to make mistakes.

f'(x) = (1/2)(10 - 3x)(5 - x)^-1/2

 

[frosty8688]

I know a lot of sites have this as the first derivative $\sqrt{5-x}(1-\frac{x}{2(5-x)})$ I am wondering what happens to the square root on the bottom? I am just wondering what is easier to work with for the second derivative.

 

[frosty8688]

Using the product rule, I have $\frac{1}{2} (10-3x)*\frac{-1}{2}(5-x)^{-3/2}*-1-3(5-x) = \frac{1}{4}\frac{10-3x-15+3x}{(5-x)^{3/2}}$ Let me know if I did anything wrong.

 

[frosty8688]

I see what I did wrong, I forgot to multiply 1/2 all the way through.