How do you differentiate $f (x) = sec (\frac{1}{\sqrt{3 x^{2} - 4}})$ $f(x)=sec(1/sqrt(3x^2-4) )$ using the chain rule?

Question

How do you differentiate $f (x) = sec (\frac{1}{\sqrt{3 x^{2} - 4}})$ $f(x)=sec(1/sqrt(3x^2-4) )$ using the chain rule?

1 Answer

Impact of this question

1561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-26T18:34:53

$\frac{d y}{d x} = - \frac{3 x}{{(3 x^{2} - 4)}^{\frac{3}{2}}} sec (\frac{1}{\sqrt{3 x^{2} - 4}}) tan (\frac{1}{\sqrt{3 x^{2} - 4}})$ $(dy)/(dx)=-(3x)/(3x^2-4)^(3/2)sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))$

Explanation:

Let ,

$y = sec u$ $color(red)(y=secu$ , $u = \frac{1}{v} and v = \sqrt{3 x^{2} - 4}$ $color(red)(u=1/v and v=sqrt(3x^2-4$

$\Rightarrow \frac{d y}{d u} = sec u tan u,$ $=>(dy)/(du)=secutanu ,$ $\frac{d u}{d v} = - \frac{1}{v^{2}} and$ $(du)/(dv)=-1/v^2 and$

$\frac{d v}{d x} = \frac{1}{2 \sqrt{3 x^{2} - 4}} \frac{d}{d x} (3 x^{2} - 4) = \frac{6 x}{2 \sqrt{3 x^{2} - 4}} = \frac{3 x}{\sqrt{3 x^{2} - 4}}$ $(dv)/(dx)=1/(2sqrt(3x^2-4))d/(dx)(3x^2-4)=(6x)/(2sqrt(3x^2-4))=(3x)/sqrt(3x^2-4)$

Using chain rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} . \frac{d v}{d x}$ $color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dv) .(dv)/(dx)$

$\frac{d y}{d x} = sec u tan u (- \frac{1}{v^{2}}) \frac{3 x}{\sqrt{3 x^{2} - 4}}, w h e r e, u = \frac{1}{v}$ $(dy)/(dx)=secutanu(-1/v^2)(3x)/sqrt(3x^2-4) ,where,u=1/v$

$\frac{d y}{d x} = sec (\frac{1}{v}) tan (\frac{1}{v}) (- \frac{1}{v^{2}}) \frac{3 x}{\sqrt{3 x^{2} - 4}}$ $(dy)/(dx)=sec(1/v)tan(1/v)(-1/v^2)(3x)/sqrt(3x^2-4)$

Putting , $v = \sqrt{3 x^{2} - 4}$ $v=sqrt(3x^2-4)$

$\frac{d y}{d x}$ $(dy)/(dx)$ = $- sec (\frac{1}{\sqrt{3 x^{2} - 4}}) tan (\frac{1}{\sqrt{3 x^{2} - 4}}) \frac{1}{3 x^{2} - 4} \cdot \frac{3 x}{\sqrt{3 x^{2} - 4}}$ $-sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))1/(3x^2-4)*(3x)/sqrt(3x^2-4)$

$\frac{d y}{d x} = - \frac{3 x}{{(3 x^{2} - 4)}^{\frac{3}{2}}} sec (\frac{1}{\sqrt{3 x^{2} - 4}}) tan (\frac{1}{\sqrt{3 x^{2} - 4}})$ $(dy)/(dx)=-(3x)/(3x^2-4)^(3/2)sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))$

Science

Math

Humanities

... and beyond

How do you differentiate $f (x) = sec (\frac{1}{\sqrt{3 x^{2} - 4}})$ $f(x)=sec(1/sqrt(3x^2-4) )$ using the chain rule?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you differentiate f(x)=sec(1√3x2−4)f(x)=sec(1/sqrt(3x^2-4) ) using the chain rule?

1 Answer

Explanation:

Related questions

Impact of this question

How do you differentiate $f (x) = sec (\frac{1}{\sqrt{3 x^{2} - 4}})$ $f(x)=sec(1/sqrt(3x^2-4) )$ using the chain rule?