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How do you use the chain rule to differentiate #y=sqrt(x^2-7x)#?

Answer 1

Samantha Adkison

#dy/dx=(2x-7)/(2sqrt(x^2-7))#

First, let #u=x^2-7x# and #y=sqrt(u)#

The chain rule uses the above manipulation in the following equation.

#dy/dx=(dy)/(du)(du)/(dx)#

Take the derivative of #u# and #y# respectively

#(dy)/(du)=d/(du)(sqrt(u))=1/(2sqrt(u))#

#(du)/(dx)=d/(dx)(x^2-7x)=2x-7#

Now multiply them, recalling that #u=x^2-7x#

#dy/dx=(dy)/(du)*(du)/(dx)#

#=1/(2sqrt(u))*(2x-7)# #=(2x-7)/(2sqrt(x^2-7))#

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Answer 2

Samantha Adkison

To differentiate ( y = \sqrt{x^2 - 7x} ) using the chain rule:

Identify the outer function ( u ) and the inner function ( v ).
- Let ( u = \sqrt{v} ) where ( v = x^2 - 7x ).
Find the derivatives of ( u ) and ( v ) with respect to ( x ).
- ( \frac{du}{dv} = \frac{1}{2\sqrt{v}} )
- ( \frac{dv}{dx} = 2x - 7 )
Apply the chain rule: ( \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} ).
- ( \frac{dy}{dx} = \frac{1}{2\sqrt{v}} \cdot (2x - 7) )
Substitute the expression for ( v ) back into the equation.
- ( \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 - 7x}} \cdot (2x - 7) )

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Answer from HIX Tutor

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