anonymous
  • anonymous
question is in my next post. please help :D
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Determine an equation for the tangent to the curve |dw:1371327715677:dw| at the point where x=-2
anonymous
  • anonymous
first derivative chain rule quotient rule
anonymous
  • anonymous
point slope formula convert to slope intercept equation

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anonymous
  • anonymous
do i separate the ^3 into the top and bottom or leave it for the end?
anonymous
  • anonymous
chain rule
anonymous
  • anonymous
you could but i think its unnecessary
anonymous
  • anonymous
ok i'm getting a slop of -512 and that is not correct cuz the text book gives me a slop of -216
anonymous
  • anonymous
First differentiate using Chain Rule and Quotient Rule:\[\bf f'(x)=\frac{6x(x^2-1)^2(4x+7)^3-12(4x+7)^2(x^2-1)^3}{(4x+7)^6}\]\[\bf =\frac{6(x^2-1)^2\cancel{(4x+7)^2}\left[ x(4x+7)-2(x^2-1) \right]}{(4x+7)^4\cancel{(4x+7)^2}}\]\[\bf =\frac{6(x^2-1)^2( 2x^2+7x+1 )}{(4x+7)^4}\]Now just fine f'(-2), that will be your slope for the tangent line. And fine f(-2) to get a point that the tangent line goes through. Use slope intercept form: \(\bf y =mx+b\) and plug in your slope for m, and the point you have for x and y then solve for b. And you're done. @0202
anonymous
  • anonymous
Note that the quadratic in the numerator of the final simplified form of the derivative should have a " +2 " in place of " +1". It was a typo. @0202
anonymous
  • anonymous
omg thank you so much!!

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