How to differentiate the function, $7x^5\sqrt{3x^4-5x^2}$

Question

$f'(x)= 7x^5(3x^4-5x^2)^{1/2}$

$= 35x^4(3x^4-5x^2)^{-1/2} +1/2(3x^4-5x^2)^{-1/2}(12x^3-10x)(7x^5)$

$= 35x^4(3x^4-5x^2)^{-1/2} + 1/2 (12x^3-10x)(7x^5)$

That is what I have done so far using the product rule and the chain rule. However, I don't know how to move forward. I also apologize for the horrible format but I really don't know how to code it nicely.

Answer 1

Given that

$f(x) = 7 x^5 \sqrt{3 x^4 - 5 x^2}$

Using the product rule, we find:

$f'(x) = 7 (5 x^4) \sqrt{3 x^4 - 5 x^2} + 7 x^5 \left( {1 \over 2 \sqrt{3 x^4 - 5 x^2}} (12 x^3 - 10 x) \right)$

Simplifying, we get

$f'(x) = 35 x^4 \sqrt{3 x^4 - 5 x^2} + {7 x^6 (6 x^2 - 5) \over \sqrt{3 x^4 - 5 x^2}}$

The answer can be further simplified as follows:

$ f'(x) = 35 x^5 \sqrt{3 x^2 - 5} + { 7 x^5 (6 x^2 - 5) \over \sqrt{3 x^2 - 5} } $

i.e.

$ f'(x) = { 35 x^5 (3 x^2 - 5) + 7 x^5 (6 x^2 - 5) \over \sqrt{3 x^2 - 5}} = {7 x^5 (15 x^2 - 25 + 6 x^2 - 5) \over \sqrt{3 x^2 - 5}} = {7 x^5 (21 x^2 - 30) \over \sqrt{3 x^2 - 5}} $

Further simplifying, we arrive at

$ f'(x) = {21 x^5 (7 x^2 - 10) \over \sqrt{3 x^2 - 5}} $

Answer 2

You can simplify it a bit as follows:

$$f(x)=7x^5\sqrt{3x^4-5x^2}=7x^6(3x^2-5)^{1/2}$$

Then

\begin{eqnarray} f'(x)&=&42x^5(3x^2-5)^{1/2}+7x^6\left(\frac{1}{2}\right)(3x^2-5)^{-1/2}(6x)\\ &=&7x^5[6(3x^2-5)^{1/2}+3x^2(3x^2-5)^{-1/2}]\\ &=&7x^5(3x^2-5)^{-1/2}[6(3x^2-5)+3x^2]\\ &=&\frac{7x^5(21x^2-30)}{\sqrt{3x^2-5}}\\ &=&\frac{21x^5(7x^2-10)}{\sqrt{3x^2-5}}\end{eqnarray}

Answer 3

Often when you have products, its beneficial to take a log first and then take derivates. So,

if $f(x)=x^5(3x^4-5x^2)^{0.5}$, then $k(x)= \log f(x)= 5\log x+0.5\log(3x^4-5x^2)$. Hence, $k'(x)= f'/f=5/x+0.5\frac{12x^3-10x}{3x^4-5x^2}$, which leads to $$f'=k'.f=x^5(3x^4-5x^2)^{0.5}\left(5/x+0.5\frac{12x^3-10x}{3x^4-5x^2}\right).$$

Answer 4

\begin{eqnarray} f(x)&=&7x^{5}\sqrt{3x^4-5x^2}\\ f(x)&=&7x^{5}\sqrt{x^2(3x^2-5)}\\ f(x)&=&7x^{5}x\sqrt{3x^2-5}\\ f(x)&=&7x^{6}\sqrt{3x^2-5}\\ \dfrac{1}{7}f^{'}(x)&=&x^6\dfrac{1}{2}\dfrac{1}{\sqrt{3x^2-5}}6x+6x^5\sqrt{3x^2-5}\\ \dfrac{1}{7\cdot6x^5}f^{'}(x)&=&\dfrac{1}{2\sqrt{3x^2-5}}x^2+\sqrt{3x^2-5}\\ \dfrac{1}{7\cdot6x^5}f^{'}(x)&=&\dfrac{x^2+2(3x^2-5)}{2\sqrt{3x^2-5}}\\ \dfrac{1}{7\cdot3x^5}f^{'}(x)&=&\dfrac{7x^2-10}{\sqrt{3x^2-5}}\\ f^{'}(x)&=&\dfrac{21x^5(7x^2-10)}{\sqrt{3x^2-5}} \end{eqnarray}

Answer 5

Yet another one (to avoid product rule): $$\begin{align} f(x)&=7\sqrt{3x^{14}-5x^{12}}\\ \text{hence}\\ f'(x)&=7\frac{42x^{13}-60x^{11}}{2\sqrt{3x^{14}-5x^{12}}}\\ &=21\frac{x^{11}(7x^2-10)}{x^6\sqrt{3x^2-5}}\\ &=21x^5\frac{7x^2-10}{\sqrt{3x^2-5}}. \end{align}$$