Find the derivative using first principles
y= 3
sqrtx
Can someone help me solve this tks
Differentiating using first principles means they want you to use the limit definition of a derivative. There are two equivalent definitions that are used, which one did you learn?
Definition 1:
For a function $\displaystyle f(x)$, it's derivative, $\displaystyle f'(x)$, at an arbitrary point $\displaystyle x$ is given by:
$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$
Defintion 2:
For a function $\displaystyle f(x)$, it's derivative, $\displaystyle f'(x)$, at an arbitrary point $\displaystyle a$ is given by:
$\displaystyle f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$
I think the first definition is what you want, so just plug the pieces into the formula and find the limit. (Do you need help with that?)
$\displaystyle f(x) = \frac {3}{ \sqrt {x}}$
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac { \frac {3}{ \sqrt {x + h}} - \frac {3}{ \sqrt {x}}}{h}$ ...........................................Combine the fractions in the numerator
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {\frac {3 \sqrt {x} - 3 \sqrt {x + h}}{\sqrt {x} \sqrt {x + h}}}{h}$ ..............................................Now combined the fractions once again
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}}$ .....................................Now multiply by the conjugate of the numerator over itself (This is a standard approach).
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}} \cdot \frac {3 \sqrt {x} + 3 \sqrt {x + h}}{3 \sqrt {x} + 3 \sqrt {x + h}}$ ...........Now simplify this
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {9x - 9(x + h)}{h \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ...........Simplify more
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {-9h}{h \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ............Now we see that the $\displaystyle h$'s cancel
$\displaystyle \Rightarrow f'(x) = \lim_{h \to 0} \frac {-9}{ \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ...............Now there's no problem when we take the limit
$\displaystyle \Rightarrow f'(x) = \frac {-9}{6 \sqrt {x} x}$
$\displaystyle \Rightarrow f'(x) = - \frac {3}{2} x^{- \frac {3}{2}}$