Evaluate the derivatives of the function. Do NOT simplify your answer. Please show me how to do this step by step. f(x)=√(16-x^2) +3arccos(x)/(4)

Question

Evaluate the derivatives of the function. Do NOT simplify your answer. Please show me how to do this step by step.


`f(x)=√(16-x^2) +3arccos(x)/(4)`

Answer 1

`f'(x)=-(x+3)/(√(16-x^2))`

Explanation:

Given function:

`f(x)=√(16-x^2)+3\arccos\left((x)/(4)\right)`

To differentiate the given function, we can use the sum rule for differentiation which states that the derivative of a sum of functions is the sum of the derivatives of those functions.

First term

Differentiate the first term
`√(16-x^2)` using the chain rule.

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

`\textsf{Let}\;\;y=√(u)\;\;\textsf{where}\;\;u=16-x^2`

Differentiate the two parts separately:

`\frac{\text{d}y}{\text{d}u}=(1)/(2)\cdot u^{(1)/(2)-1}=(1)/(2) \cdot u^{-(1)/(2)}=(1)/(2√(u))`

`\frac{\text{d}u}{\text{d}x}=0-2\cdot x^(2-1)=-2x`

Put everything into the chain rule formula:

`\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}* \frac{\text{d}u}{\text{d}x}`

`\frac{\text{d}y}{\text{d}x}=(1)/(2√(u))* (-2x)`

`\frac{\text{d}y}{\text{d}x}=(-2x)/(2√(u))`

`\frac{\text{d}y}{\text{d}x}=-(x)/(√(u))`

Substitute back
`u=16-x^2`:

`\frac{\text{d}y}{\text{d}x}=-(x)/(√(16-x^2))`

Second term

Differentiate the second term using the chain rule and the differentiation rule for arccos(x):

`\boxed{\frac{\text{d}}{\text{d}x}(\arccos x)=-(1)/(√(1-x^2))}`

`\textsf{Let}\;\;y=3\arccos(u)\;\;\textsf{where}\;\;u=(x)/(4)`

Differentiate the two parts separately:

`\frac{\text{d}y}{\text{d}u}=-(3)/(√(1-u^2))`

`\frac{\text{d}u}{\text{d}x}=(1)/(4)`

Put everything into the chain rule formula:

`\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}* \frac{\text{d}u}{\text{d}x}`

`\frac{\text{d}y}{\text{d}x}=-(3)/(√(1-u^2))* (1)/(4)`

`\frac{\text{d}y}{\text{d}x}=-(3)/(4√(1-u^2))`

Substitute back u = x/4:

`\frac{\text{d}y}{\text{d}x}=-\frac{3}{4\sqrt{1-\left((x)/(4)\right)^2}}`

Rewrite 4 as √(16):

`\frac{\text{d}y}{\text{d}x}=-\frac{3}{√(16)\sqrt{1-\left((x)/(4)\right)^2}}`

`\textsf{Apply the radical rule:} \quad \sqrt{\vphantom{b}a}√(b)=√(ab)`

`\frac{\text{d}y}{\text{d}x}=-\frac{3}{\sqrt{16\left(1-\left((x)/(4)\right)^2\right)}}`

`\frac{\text{d}y}{\text{d}x}=-\frac{3}{\sqrt{16-16\left((x)/(4)\right)^2}}`

`\frac{\text{d}y}{\text{d}x}=-\frac{3}{\sqrt{16-16\cdot(x^2)/(16)}}`

`\frac{\text{d}y}{\text{d}x}=-(3)/(√(16-x^2))`

Now, we have the derivatives of both terms, and we can add them together to find the derivative of f(x):

`f'(x)=-(x)/(√(16-x^2))-(3)/(√(16-x^2))`

As the denominators of both terms are the same, we can combine them:

`f'(x)=(-x-3)/(√(16-x^2))`

`f'(x)=-(x+3)/(√(16-x^2))`

Therefore, the derivative of the given function is:

`\large\boxed{\boxed{f'(x)=-(x+3)/(√(16-x^2))}}`