Chain Rule Problem with multiple square roots

[sarahjohnson]

Find the derivative of the function.

y =

7x + 7x + 7x

$\displaystyle \displaystyle y \ = \ \sqrt {7x \ + \ \sqrt {7x \ + \ \sqrt {7x}}}$

sqrt(7x +sqrt (7x +sqrt 7x))

I got 1/2(7x+(7x+(7x)^1/2)^1/2)^-1/2 (7+1/2(7x+(7x)^1/2)^1/2) (7+1/2(7x)^-1/2) (7)

sorry for the long and confusing way it's typed out. I'm not sure how to use symbols on here.

I double checked it and I think it is right but my hw program tells me it's wrong

Thanks in advance!

 

[pka]

Find the derivative of the function.

y =

7x + 7x + 7x

$\displaystyle \displaystyle y \ = \ \sqrt {7x \ + \ \sqrt {7x \ + \ \sqrt {7x}}}$

sqrt(7x +sqrt (7x +sqrt 7x))

I got 1/2(7x+(7x+(7x)^1/2)^1/2)^-1/2 (7+1/2(7x+(7x)^1/2)^1/2) (7+1/2(7x)^-1/2) (7)

sorry for the long and confusing way it's typed out. I'm not sure how to use symbols on here.

I agree that it is confusing. I don't really follow it.

Look at this webpage.

 

[Deleted member 4993]

Find the derivative of the function.

y =

7x + 7x + 7x

$\displaystyle \displaystyle y \ = \ \sqrt {7x \ + \ \sqrt {7x \ + \ \sqrt {7x}}}$

sqrt(7x +sqrt (7x +sqrt 7x))

I got 1/2(7x+(7x+(7x)^1/2)^1/2)^-1/2 (7+1/2(7x+(7x)^1/2)^1/2) (7+1/2(7x)^-1/2) (7)

sorry for the long and confusing way it's typed out. I'm not sure how to use symbols on here.

I double checked it and I think it is right but my hw program tells me it's wrong

Thanks in advance!

I get

$\displaystyle \displaystyle y' \ = \ \frac{1}{2}\left [7x + (7x + \sqrt{7x})^{\frac{1}{2}}\right ] ^{-\frac {1}{2}} \ * \$

...........$\displaystyle \displaystyle \left [ 7 \ + \ \frac{1}{2} \ * \ \left (7x+\sqrt{7x}\right )^{-\frac{1}{2}} \ * \left (7 \ + \ \frac{\sqrt 7}{2\sqrt x}\right )\right ]$

....

 

[mathmari]

$\displaystyle \left( \sqrt{7x + \sqrt{7x +\sqrt{7x}}}\right)' \, = \, \dfrac{7+\left( \sqrt {7x + \sqrt {7x}}\right)'}{2 \sqrt{7x +\sqrt {7x + \sqrt {7x}}}}$

. . . . .$\displaystyle = \, \dfrac{7+\dfrac{7+\left(\sqrt{7x}\right)'}{2 \sqrt{7x+ \sqrt{7x}}}}{2 \sqrt{7x + \sqrt {7x + \sqrt {7x}}}}\, = \,\dfrac{7+\dfrac{7+\dfrac{7}{2\sqrt{7x}}}{2 \sqrt{7x+ \sqrt{7x}}}}{2 \sqrt{7x + \sqrt {7x + \sqrt {7x}}}}$

 

[sarahjohnson]

thank you guys! Finally got it. You guys are awesome