How do you find the derivative of ${(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}$ $(3x-(sin4x)^2)^(1/2)$ ?

Question

How do you find the derivative of ${(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}$ $(3x-(sin4x)^2)^(1/2)$ ?

1 Answer

Impact of this question

2196 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T10:10:29

See solution below.

Explanation:

$\frac{d}{d x} [{(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}]$ $d/dx[(3x-(sin4x)^2)^{1/2}]$

This one is a bit of a hassle but with proper substitution it's possible. Think of this expression as multiple functions.
${(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}$ $(3x-(sin4x)^2)^{1/2}$ is a function of $f$ $f$ where $f (x) = x^{\frac{1}{2}}$ $f(x)=x^{1/2}$ and $x = 3 x - {(sin 4 x)}^{2}$ $x=3x-(sin4x)^2$ .
Always remember to multiply the derivative of the inner function by the derivative of the outer function. It's a bit hard for me to explain so please follow the process below.

$\frac{d}{d x} [{(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}]$ $d/dx[(3x-(sin4x)^2)^{1/2}]$

Start by using the substitution: $u = 3 x - {(sin 4 x)}^{2}$ $u=3x-(sin4x)^2$

$\frac{d}{d x} [{(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}] = \frac{d}{d u} [u^{\frac{1}{2}}] \times \frac{d}{d x} [3 x - {(sin 4 x)}^{2}]$ $d/dx[(3x-(sin4x)^2)^{1/2}]=d/{du}[u^{1/2}] times d/dx[3x-(sin4x)^2]$

$\frac{d}{d x} [3 x - {(sin 4 x)}^{2}] = \frac{d}{d x} [3 x] - \frac{d}{d x} [{(sin 4 x)}^{2}]$ $d/dx[3x-(sin4x)^2]= d/dx[3x]-d/dx[(sin4x)^2]$

$v = sin 4 x$ $v=sin4x$

$\frac{d}{d x} [{(sin 4 x)}^{2}] = \frac{d}{d v} [v^{2}] \frac{d}{d x} [sin 4 x]$ $d/dx[(sin4x)^2] = d/{dv}[v^2]d/dx[sin4x]$

$w = 4 x$ $w=4x$

$\frac{d}{d x} [sin 4 x] = \frac{d}{d w} [sin w] \frac{d}{d x} [4 x]$ $d/dx[sin4x]=d/{dw}[sinw]d/dx[4x]$

Now putting all this together we get

$\frac{d}{d u} [u^{\frac{1}{2}}] \times (\frac{d}{d x} [3 x] - (\frac{d}{d v} [v^{2}] (\frac{d}{d w} [sin w] \frac{d}{d x} [4 x])))$ $d/{du}[u^{1/2}] times (d/dx[3x]-(d/{dv}[v^2] (d/{dw}[sinw]d/dx[4x])))$

evaluating this expression we get:

$\frac{1}{2} u^{- \frac{1}{2}} \times (3 - (2 v (cos w \times 4)))$ $1/2 u^{-1/2} times (3-(2v (cosw times 4)))$

Substituting all the values we get:

$\frac{1}{2} {(3 x - {(sin 4 x)}^{2})}^{- \frac{1}{2}} \times (3 - (2 (sin 4 x) (cos (4 x) \times 4)))$ $1/2 (3x-(sin4x)^2)^{-1/2} times (3-(2(sin4x) (cos(4x) times 4)))$

This can be simplified to:

$\frac{3 - 8 sin 4 x cos 4 x}{2 \times \sqrt{(3 x - {sin}^{2} (4 x))}}$ $(3-8sin4xcos4x)/{2 times sqrt{(3x-sin^2(4x))}$

Science

Math

Humanities

... and beyond

How do you find the derivative of ${(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}$ $(3x-(sin4x)^2)^(1/2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of (3x−(sin4x)2)12(3x-(sin4x)^2)^(1/2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the derivative of ${(3 x - {(sin 4 x)}^{2})}^{\frac{1}{2}}$ $(3x-(sin4x)^2)^(1/2)$ ?