Question

Given f(x)=x^2 √x . Find f'(x) using the definition of derivative. can anyone help me with this ?

Answer 1

\[f(x)=x^2*x^1/2=x^{2+1/2}=x^{5/2}\]
\[f'(x)= \frac{5}{2}*x^{5/2-1}\]
=5/2*x^3/2

Answer 2

\[= \frac{5}{2}x^ \frac{3}{2}\]

Answer 3

The conventional definition of a derivative is via a difference quotient, of the form
\[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \]
In this case, that means
\[f'(x) = \lim_{h \rightarrow 0} \frac{(x+h)^2\sqrt{x+h} - x^2\sqrt{x}}{h} \]

Answer 4

jemurray3, what should i do after that ?

Answer 5

I would have left it as\[x^2\sqrt x=x^{5/2}\]then the numerator can expand and the only the second term in the series will survive after the limit is taken, but I want to see Jemurray's way...

Answer 6

ok, no reply from Jemurray so I'll go...

Answer 7

or not.

Answer 8

haha I just accidentally deleted my thing. Damn. Mine is algebraically more involved but I assumed expansions were out of the question at this level. Give me jussssst a second...

Answer 9

I keep word open and copy-past when I'm using a lot of LaTeX for that very reason.

Answer 10

After some simplification, the numerator becomes
\[x^2(\sqrt{x+h} -\sqrt{x}) + h^2\sqrt{x+h} + 2xh\sqrt{x+h} \]
Dividing by h and taking the limit, we are left with
\[x^2\left( \lim_{h\rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \right) + 2x\sqrt{x} \]
The thing in parentheses is the derivative of sqrt(x), which we can evaluate as follows:
\[\lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}= \lim_{h \rightarrow 0 }\frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}\]
Plugging that in, we get
\[\frac{x^2}{2\sqrt{x}} + 2x\sqrt{x} = \frac{x^{\frac{3}{2}}}{2} + 2x^{\frac{3}{2}} = \frac{5}{2} x^{\frac{3}{2}}\]
for our derivative. Whew.

Answer 11

@TuringTest good call, I'll do that from now on!

Answer 12

Learned the hard way.

I'm not going to finish, yours is similar enough

Answer 13

Thank you. :)