Finding the derivative of a function by limit definition

Question

I want to know how to find the derivative of functions of this sort by definition: $f(x)=\frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt[5]{x}+\sqrt[4]{x^5}}.$

Specifically this one, I've been struggling with. How does one approach these kinds of limits?

$$f'(x)=\displaystyle{\lim_{h \to 0}}\left(\frac{\frac{\sqrt{x+h}+\sqrt[3]{x+h}}{\sqrt[5]{x+h}+\sqrt[4]{(x+h)^5}}-\frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt[5]{x}+\sqrt[4]{x^5}}}{h}\right).$$

I've been trying for a while now. It just seems impossible. Thanks in advance!

Answer 1

While the quotient rule is certainly valid, I have found that it is easier to stick with the product rule. I would also remove all of the radical signs. I would therefore re-write the function as

$$f(x) = \left[\left(x\right)^{(1/2)} + \left(x\right)^{(1/3)}\right] ~\times~ \left[\left(x\right)^{(1/5)} + \left(x\right)^{(4/5)}\right]^{(-1)}. \tag1$$

Then, I would (again) re-write the function as $$f(x) = A(x) \times \left[B(x)\right]^{(-1)}\tag2 $$

where

$$A(x) = \left[\left(x\right)^{(1/2)} + \left(x\right)^{(1/3)}\right] ~~\text{and}~~ B(x) = \left[\left(x\right)^{(1/5)} + \left(x\right)^{(4/5)}\right]. \tag3 $$

At this point, I would focus on equation (2) above, noting that

$$f'(x) = A(x) ~\left\{(-1)\left[B(x)\right]^{(-2)} B~'(x)\right\} ~+~ \left[B(x)\right]^{(-1)}A'(x). \tag4$$

At this point, it remains to :

• manually compute $A'(x)$ and $B'(x)$.
• use the values for $A(x), B(x), A'(x),$ and $B'(x)$ to fill-in-the-blanks for equation (4).

$$ A'(x) = (1/2)~x^{-(1/2)} ~+~ (1/3)~x^{-(2/3)}. \tag5$$

$$ B'(x) = (1/5)~x^{-(4/5)} ~+~ (4/5)~x^{-(1/5)}. \tag6$$

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So now, it only remains to combine the values from equations (3), (5), and (6), in accordance with equation (4) above.

$$f'(x) = \left[\left(x\right)^{(1/2)} + \left(x\right)^{(1/3)}\right] $$

$$\times ~\left\{(-1)\left[\left(x\right)^{(1/5)} + \left(x\right)^{(4/5)}\right]^{(-2)} \left[(1/5)~x^{-(4/5)} ~+~ (4/5)~x^{-(1/5)}\right]\right\}$$

$$ ~+~ \left[\left(x\right)^{(1/5)} + \left(x\right)^{(4/5)}\right]^{(-1)}$$ $$\times \left[(1/2)~x^{-(1/2)} ~+~ (1/3)~x^{-(2/3)}\right]. $$

Answer 2

If $f'(x)=\displaystyle{\lim_{h \to 0}}\left(\frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\right)$

then we can write

$f’(x)=\displaystyle{\lim_{h \to 0}}\left(\frac{u(x+h)-u(x)}{h.v(x+h)}-\frac{u(x)}{v(x).v(x+h)}.\frac{v(x+h)-v(x)}{h}\right)$

Setting

$u(x)=\sqrt{x}+\sqrt[3]{x}$

and

$v(x)= \sqrt[5]{x}+x.\sqrt[4]{x}$,

we have

$u’(x)=\displaystyle{\lim_{h \to 0}}\frac{u(x+h)-u(x)}{h}=$ $=\displaystyle{\lim_{h \to 0}}\frac{\sqrt{x+h}+\sqrt[3]{x+h}-\sqrt{x}-\sqrt[3]{x}}{h}=$

$=\displaystyle{\lim_{h \to 0}}\frac{\sqrt{x+h}-\sqrt{x}}{h}.\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}+\displaystyle{ \lim_{h \to 0}}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}.\frac{\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}}}{\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}}}=$

$=\displaystyle{\lim_{h \to 0}}\frac{x+h-x}{h. (\sqrt{x+h}+\sqrt{x}) }+\displaystyle{\lim_{h \to 0}}\frac{x+h-x}{h.(\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}})}=$

$=\frac{1}{2\sqrt{x}}+\frac{1}{3\sqrt[3]{x^{2}}}.$

Moreover we have

$v’(x)=\displaystyle{ \lim_{h \to 0}}\frac{v(x+h)-v(x)}{h}=$,

$=\displaystyle{\lim_{h \to 0}}\frac{\sqrt[5]{(x+h)}+(x+h)\sqrt[4]{x+h}-\sqrt[5]{x}-x\sqrt[4]{x}}{h}=$,

$=\displaystyle{\lim_{h \to 0}}\frac{ \sqrt[5]{(x+h)}-\sqrt[5]{x}} {h}.$

$.\frac{ \sqrt[5]{(x+h)^{4}}+\sqrt[5]{x.(x+h)^{3}}+\sqrt[5]{(x+h)^{2}.x^{2}}+\sqrt[5]{(x+h).x^{3}}+\sqrt[5]{x^{4}} } { \sqrt[5]{(x+h)^{4}}+\sqrt[5]{(x+h)^{3}x}+\sqrt[5]{(x+h)^{2}x^{2}}+\sqrt[5]{(x+h)x^{3}}+\sqrt[5]{x^{4}} }$

$+ \displaystyle{\lim_{h \to 0}}\frac{(x)(\sqrt[4]{x+h}-\sqrt[4]{x}) }{h}.\frac{\sqrt[4]{(x+h)^{3}}+\sqrt[4]{(x+h)^{2}x}+\sqrt[4]{(x+h)x^{2}}+\sqrt[4]{x^{3}} }{\sqrt[4]{(x+h)^{3}}+\sqrt[4]{(x+h)^{2}x}+\sqrt[4]{(x+h)x^{2}}+\sqrt[4]{x^{3}} }+\displaystyle{\lim_{h \to 0}}\frac{ h\sqrt[4]{x+h}}{h}=$

$=\displaystyle{\lim_{h \to 0}}\frac{x+h-x}{h.(\sqrt[5]{(x+h)^{4}}+\sqrt[5]{(x+h)^{3}x}+\sqrt[5]{(x+h)^{2}x^{2}}+\sqrt[5]{(x+h)x^{3}}+\sqrt[5]{x^{4}})}+ \displaystyle{\lim_{h \to 0}}\frac{(x)(x+h-x) }{h.(\sqrt[4]{(x+h)^{3}}+\sqrt[4]{(x+h)^{2}x}+\sqrt[4]{(x+h)x^{2}}+\sqrt[4]{x^{3}}) }+\displaystyle{\lim_{h \to 0}}\frac{ h\sqrt[4]{x+h}}{h}=$

$=\frac{1}{5\sqrt[5]{x^{4}}}+\frac{x}{4\sqrt[4]{x^{3}}}+\sqrt[4]{x}$

Finally $f’(x)=\frac{8+18\sqrt[6]{x}-55x\sqrt[20]{x}-45x\sqrt[60]{x^{13}}} {60\sqrt[15]{x^{13}}(1+x\sqrt[20]{x})^{2} }.$