Question

Can someone help me solve this? f(x)=arccos(2x+4)−sqrt[3-3x^2] find f′(x)?

Answer 1

`(-2)/(sqrt(1-4x^2-16x-16)) +(3x)/(sqrt(3-3x^2))`

Explanation:

So we can differentiate each half of this equation.

We know chain rule, power rule, and the derivative of arccosine which is
`d/dx arccos(x) = -1/(sqrt(1-x^2))`

From this, we see those, we see the first term has derivative
`d/dx arccos(2x+4) = d/dx(2x+4) cdot -1/(sqrt(1-(2x+4)^2)) = (-2)/(sqrt(1-4x^2-16x-16))`

The second term has
`d/dx sqrt(3-3x^2) = d/dx (3-3x^2)^(1/2) = 1/2 cdot d/dx(3-3x^2) cdot (3-3x^2)^(-1/2) = (-3x)/(sqrt(3-3x^2))`

So the final answer is the difference between the two, as written above. There seems to be no simplification by combining the terms, which I have written out below as proof:
`3-3x^2 = 3(1-x^2) = 3(1-x)(1+x)`
`4x^2 + 16x + 15 = (2x+5)(2x+3)`

therefore,

`f'(x) = (-2)/(sqrt(1-4x^2-16x-16)) +(3x)/(sqrt(3-3x^2))`

`= (3xsqrt(-(2x+5)(2x+3)) -2sqrt(3(1-x)(1+x)) )/(sqrt(-3(1-x)(1+x)(2x+5)(2x+3)))`

Answer 2

`f'(x)=(-2sqrt(1-(2x+4)^2))/(1-(2x+4)^2)+(xsqrt(3-3x^2))/(1-x^2)`

Explanation:

.

`f(x)=arccos(2x+4)-sqrt(3-3x^2)`

`y_1=arccos(2x+4)`

`y_2=sqrt(3-3x^2)`

`y=y_1-y_2`

`dy/dx=dy_1/dx-dy_2/dx` `color(red)(Equation 1)`

`y_1=arccos(2x+4)` can be written as:

`cosy_1=2x+4`

Let's take derivatives of both sides:

`-siny_1dy_1=2dx`

Let's divide both sides by `-siny_1dx`:

`dy_1/dx=-2/siny_1`

But we know:

`sin^2y_1+cos^2y_1=1`. If we solve for `siny_1` we get:

`siny_1=sqrt(1-cos^2y_1)`

But `cos^2y_1=(2x+4)^2`. Let's plug this in:

`siny_1=sqrt(1-(2x+4)^2)`, therefore:

`dy_1/dx=-2/sqrt(1-(2x+4)^2)=(-2sqrt(1-(2x+4)^2))/(1-(2x+4)^2)`

`color(red)(dy_1/dx=(-2sqrt(1-(2x+4)^2))/(1-(2x+4)^2))`

`y_2=sqrt(3-3x^2)=(3-3x^2)^(1/2`

Let's let `u=3-3x^2`, then:

`(du)/dx=-6x`

`y_2=u^(1/2`

`dy_2/(du)=1/2u^(-1/2)`

The Chain Rule says:

`dy_2/dx=dy_2/(du)*(du)/dx`

`dy_2/dx=1/2u^(-1/2)(-6x)`

Let's substitute back for `u`:

`dy_2/dx=1/2(3-3x^2)^(-1/2)(-6x)`

`dy_2/dx=(-3x)/sqrt(3-3x^2)=((-3x)(sqrt(3-3x^2)))/(3-3x^2)`

`color(red)(dy_2/dx=(-xsqrt(3-3x^2))/(1-x^2))`

According to `color(red)(Equation 1)` above:

`dy/dx=(-2sqrt(1-(2x+4)^2))/(1-(2x+4)^2)+(xsqrt(3-3x^2))/(1-x^2)`