Question

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Please evaluate the Derivative using the Limit Process. Show all workings.

Answer 1

`\displaystyle f'(x)=-\frac{2}{x^{{}^(3)\!/\!{}_(2)}}`

Explanation:

We have the function:

`\displaystyle f(x)=(4)/(\sqrt x)`

And we want to find the derivative using the limit process.

Recall that the definition of a derivative is:

`\displaystyle \lim_(h \to 0) (f(x+h)-f(x))/(h)`

Therefore, by substitution:

`\displaystyle \lim_(h \to 0)((4)/(√(x+h))-(4)/(\sqrt x))/(h)`

First and foremost, we can move the constant factor outside of the limit:

`\displaystyle =\lim_(h \to 0)(4\left((1)/(√(x+h))-(1)/(\sqrt x)\right))/(h)\\ \\=4\lim_(h \to 0)((1)/(√(x+h))-(1)/(\sqrt x))/(h)`

Next, we can multiply everything by (√(x + h)(√x) to eliminate the fractions in the denominator. Therefore:

`\displaystyle =4\lim_(h \to 0)((1)/(√(x+h))-(1)/(\sqrt x))/(h)\left((√(x+h)\sqrt x)/(√(x+h)\sqrt x)\right)`

Distribute:

`\displaystyle =4\lim_(h \to 0)\frac{\left({√(x+h)\sqrt x}\right)(1)/(√(x+h))-(√(x+h)\sqrt x)(1)/(\sqrt x)}{h({√(x+h)\sqrt x})}`

Distribute and simplify:

`\displaystyle =4 \lim_(h\to 0)(\sqrt x-√(x+h))/(h(√(x+h)√(x)) )`

Next, we can multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x + h)). Thus:

`\displaystyle = 4\lim_(h\to 0)(\sqrt x-√(x+h))/(h(√(x+h)√(x)) )\left((\sqrt x +√(x+h))/(\sqrt x +√(x+h)\right))`

Simplify:

`\displaystyle =4 \lim_(h \to 0) (x-(x+h))/(h(√(x+h)\sqrt x)(\sqrt x+√(x+h)))\\ \\ \\ =4 \lim_(h \to 0) (x-x-h)/(h(√(x+h)\sqrt x)(\sqrt x+√(x+h))) \\ \\ \\=4 \lim_(h \to 0) (-h)/(h(√(x+h)\sqrt x)(\sqrt x+√(x+h)))`

Cancel like terms:

`\displaystyle =4 \lim_(h \to 0) -(1)/((√(x+h)\sqrt x)(\sqrt x+√(x+h)))`

Now, we can use direct substitution. Hence:

`\displaystyle \Rightarrow4 \left( -(1)/((√(x+0)\sqrt x)(\sqrt x+√(x+0)))\right)`

Simplify:

`\displaystyle =4\left( -(1)/((√(x)\sqrt x)(\sqrt x+√(x)))\right) \\ \\ \\ =4\left( -(1)/((x)(2√(x)))\right)`

Multiply:

`\displaystyle =- (4)/(2x√(x))`

Reduce and rewrite:

`\displaystyle =-\frac{2}{x(x^{{}^(1)\! / \! {}_(2) \!})}`

Simplify:

`\displaystyle =-\frac{2}{x^{{}^(3)\!/\!{}_(2)}}`

Therefore:

`\displaystyle f'(x)=-\frac{2}{x^{{}^(3)\!/\!{}_(2)}}`