How do you find the derivative of the function: #sqrt(1-9x^2)arccos(3x)#?

Question

Evelyn Agard · Accepted Answer

#y=sqrt(1-9x^2)arccos(3x)#

Let #" "3x=costheta#

So Differentiating w r to x we have #" " 3=-sintheta(d theta)/(dx)#

#=>(d theta)/(dx)=-3/sintheta#

Now

#y=sqrt(1-9x^2)arccos(3x)#

#=>y=sqrt(1-cos^2theta)arccos(costheta)#

#=>y=thetasintheta#

Differentiating w r to #theta# we have

#(dy)/(d theta)=thetacostheta+sintheta#

So

#(dy)/(dx)=(dy)/(d theta)xx(d theta)/(dx)#

#=>(dy)/(dx)=(thetacostheta+sintheta)xx(-3/sintheta)#

#=>(dy)/(dx)=(-3thetacostheta/sintheta-3)#

#=>(dy)/(dx)=-3((thetacostheta)/sqrt(1-cos^2theta)+1)#

#=>(dy)/(dx)=-3((3xcos^-1(3x))/sqrt(1-9x^2)+1)#

William Abdalla · Answer

undefined To find the derivative of the function $ \sqrt{1-9x^2} \cdot \text{arccos}(3x) $, you can use the product rule of differentiation.

The product rule states that if you have two functions $ u(x) $ and $ v(x) $, then the derivative of their product $ u(x)v(x) $ with respect to $ x $ is given by $ u'(x)v(x) + u(x)v'(x) $.

Let $ u(x) = \sqrt{1-9x^2} $ and $ v(x) = \text{arccos}(3x) $.

Then, find the derivatives $ u'(x) $ and $ v'(x) $, and apply the product rule to get the derivative of the given function.

Here are the steps:

1. Find $ u'(x) $:
$$ u'(x) = \frac{d}{dx}(\sqrt{1-9x^2}) $$

2. Find $ v'(x) $:
$$ v'(x) = \frac{d}{dx}(\text{arccos}(3x)) $$

3. Apply the product rule:
$$ \frac{d}{dx}(\sqrt{1-9x^2} \cdot \text{arccos}(3x)) = u'(x)v(x) + u(x)v'(x) $$