# Let f(x)=sqrt(1-sinx)What is the domain of f(x), find f'(x), what is the domain of f'(x) and write an equation for the line tangent to the graph of f at x=0

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### 3 Answers

`f(x) = sqrt(1-sin(x))`

Domain of f(x) -->

Domain of sin(x) is all the Real numbers, `x in R` . For all the real numbers, sine would be in the following range,

`-1lt=sin(x)lt=+1`

Therefore, multiplying by -1,

`+1=gt-sin(x)=gt-1`

Adding +1,

`1+1=gt1-sin(x)=gt1-1`

Therefore,

`2=gt1-sin(x)=gt0`

We know for all real numbers `1- sin(x)` will be positive.

Therefore the expression inside the squareroot is always positive for all `x in R` . **Therefore the Domain of f(x) is,**

**Domain of** f`(x) = {x in R}`

`f'(x) = (1/2)(-1)(cos(x))(1-sin(x))^(-1/2)`

`f'(x) = (-cos(x))/(2sqrt(1-sin(x)))`

Now since we found the domain and range of 1-sin(x) above, finding the domain of this easy.

The only concern here is the denominator. The denominator cannot be zero. That means,

`1-sin(x) != 0`

`sin(x) != 1`

x not equal to **`npi+(-1)^n(pi/2)` ** where n is any integer.

so x` != {1/2pi, 5/2pi and 9/2pi...}`

or `x!= {1/2pi, 2pi+1/2pi and 4pi+1/2pi...}`

**Therefore domain of f'(x) is given by,**

**Domain of** `f'(x) = {x in R, x != 2npi+1/2pi, n in Z}`

To find the line tangent to f(x) at x =0,

we have to find f'(x) at x = 0.

`f'(0) = (-cos(0))/(2sqrt(1-sin(0)))`

`f'(0) = (-1)/(2sqrt(1-0))`

`f'(0) = -1/2`

This is the gradient of the line,

and we need f(0),

`f(0) = sqrt(1-sin(0))`

`f(0) = sqrt(1-0)`

`f(0) = sqrt(1-sin(x))`

`f(0) = 1`

This is the intercept of the line,

**Therefore the equation is,**

`y = mx+c`

`y = -1/2x+1`

- Domain - that's the set of numbers we can put in. We have a square root, which must be of a positive number, so:

1-sinx >= 0

So sinx < 1. Phew! x can be anything, because sinx is ALWAYS 1 or less.

- Derivative - you'll need the *chain rule*:

(d/dx) u(v) = du/dv * dv/dx

v = 1-sinx

u = sqrt(x)

So f'(x) = (d/dx) sqrt(1-sinx) = (d/dv) (1-v)^(1/2) * (d/dx) (1-sinx)

Let's tackle the second bit first. The derivative of a constant is zero, and the derivative of -sinx is -cosx, so (d/dx)(1-sinx) = -cosx.

We'll have to apply the chain rule to the first bit again:

(d/dv) (1-v)^(1/2) = (d/da) (a^(1/2)) * (d/dv) (1-v)

a = 1-v

The first bit: standard differentiation - power in front, take 1 away from the power:

(d/da) (a^(1/2)) = (1/2) a^(-1/2) = 1/(2 * sqrt(a)) = 1/(2 * sqrt(1-sinx))

Second bit: constant differentiates to zero, -v differentiates to -1.

So (d/dv) (1-v)^(1/2) = 1/(2 * sqrt(1-sinx)) * -1

And f'(x) = (d/dx) sqrt(1-sinx) = (d/dv) (1-v)^(1/2) * (d/dx) (1-sinx) = -1/(2 * sqrt(1-sinx)) * (-cosx) = cosx / (2*sqrt(1-sinx)). Done!

The domain comes from the same procedure as for f(x).

- tangent - the line that 'just touches' the graph of f at a certain point (here, it's x=0).

A tangent has the equation y=mx+c. m is the gradient, which is the derivative of f *evaluated at the point we're at*, i.e. f'(0). I'll leave you to do that, since the expression for f'(x) is above :)

c is the value of the function when x is zero. As in, f(0). Yes? :)

So you have m and c, which gives you an equation for the tangent!

i have the same question but i have to find the equation of the tangent line where x=pi. can someone please help me?