Answer to Question #96870 in Calculus for Ojugbele Daniel

Question

Differentiate from first principle square root of (x+√x).

Answer 1

$$f(x)=\sqrt{x+\sqrt{x}}$$

$$f'(x)=\bigg(\sqrt{x+\sqrt{x}}\bigg)'=\lim\limits_{h\rarr0}{f(x+h)-f(x) \over h}=$$

$$=\lim\limits_{h\rarr0}{\sqrt{x+h+\sqrt{x+h}}-\sqrt{x+\sqrt{x}} \over h}=$$

$$=\lim\limits_{h\rarr0}{\sqrt{x+h+\sqrt{x+h}}-\sqrt{x+\sqrt{x}} \over h}\cdot{\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}} \over \sqrt{x+h+\sqrt{x+h}}}=$$

$$=\lim\limits_{h\rarr0}{ x+h+\sqrt{x+h}-x-\sqrt{x}\over h(\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})}=$$

$$=\lim\limits_{h\rarr0}{ h\over h(\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})}+$$

$$+\lim\limits_{h\rarr0}{ \sqrt{x+h}-\sqrt{x}\over h(\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})}=$$

$$={ 1\over \sqrt{x+0+\sqrt{x+0}}+\sqrt{x+\sqrt{x}}}+$$

$$+\lim\limits_{h\rarr0}{ \sqrt{x+h}-\sqrt{x}\over h(\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})}\cdot{\sqrt{x+h}+\sqrt{x} \over \sqrt{x+h}+\sqrt{x}}=$$

$$={ 1\over 2\sqrt{x+\sqrt{x}}}+\lim\limits_{h\rarr0}{ x+h-x\over h(\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})(\sqrt{x+h}+\sqrt{x})}=$$

$$={ 1\over 2\sqrt{x+\sqrt{x}}}+\lim\limits_{h\rarr0}{ 1\over (\sqrt{x+h+\sqrt{x+h}}+\sqrt{x+\sqrt{x}})(\sqrt{x+h}+\sqrt{x})}=$$

$$={ 1\over 2\sqrt{x+\sqrt{x}}}+{ 1\over (\sqrt{x+0+\sqrt{x+0}}+\sqrt{x+\sqrt{x}})(\sqrt{x+0}+\sqrt{x})}=$$

$$={ 1\over 2\sqrt{x+\sqrt{x}}}\big(1+{1 \over 2\sqrt{x}}\big)$$

$$\bigg(\sqrt{x+\sqrt{x}}\bigg)'={ 1\over 2\sqrt{x+\sqrt{x}}}\bigg(1+{1 \over 2\sqrt{x}}\bigg)$$