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onegirl

  • one year ago

Find an equation of the tangent line to h(x) = f(x)g(x) at x = 1

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  1. ingenuus
    • one year ago
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    \[y=[f'(1)g(1) + f(1)g'(1)](x-1) + f(1)g(1)\]

  2. campbell_st
    • one year ago
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    well you need the point of the curve which will be (1, h(1)) you need to find the slope... but 1st you need to differentiate the function using the product rule h'(x) = f(x)g'(x) + g(x)f'(x) the slope at x = 1 is found by substituting x = 1 into the 1st derivative(above) h'(1) = f(1)g'(1) + g(1)f'(1) now you have a slope h'(1) and point (1, h(1)) use the point slope formula to find the equation of the tangent.

  3. onegirl
    • one year ago
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    ok

  4. onegirl
    • one year ago
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    so i got y = g(1)f'(1) + f(1)x g'(1) - f(1) g'(1) + f(1)g(1) is it correct?

  5. onegirl
    • one year ago
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    @campell_st ?

  6. onegirl
    • one year ago
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    @campbell_st

  7. onegirl
    • one year ago
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    @ingenuus did i get it right?

  8. ingenuus
    • one year ago
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    no the answer is above

  9. onegirl
    • one year ago
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    okay

  10. ingenuus
    • one year ago
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    do you understand the equation y=m(x-a) +b ?

  11. onegirl
    • one year ago
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    yes

  12. ingenuus
    • one year ago
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    what is the slope m in your case

  13. onegirl
    • one year ago
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    1?

  14. ingenuus
    • one year ago
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    why is it 1?

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