how to derive (10^x + 10^-x)^2?

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  • 1 month ago

    This needs a chain rule.

    y = (10^x + 10^(-x))²

    If we set this up as a chain rule we get:

    y = u² and u = 10^x + 10^(-x)

    we need the derivative of both so we get:

    dy/du = 2u and du/dx = 10^x ln(10) - 10^(-x) ln(10)

    Chain rule is:

    dy/dx = dy/du * du/dx

    substitute and simplify:

    dy/dx = 2u * [10^x ln(10) - 10^(-x) ln(10)]

    Substitute the u for the expression in terms of x:

    dy/dx = 2[10^x + 10^(-x)] * [10^x ln(10) - 10^(-x) ln(10)]

    If we factor out the ln(10) we get:

    dy/dx = 2 ln(10) [10^x + 10^(-x)] * [10^x - 10^(-x)]

    Now we can multiply the two binomials as they are conjugate pairs we should get a 2-term result:

    dy/dx = 2 ln(10) [(10^x)² - 10^x * 10^(-x) + 10^x * 10^(-x) - 10^(-x)²]

    dy/dx = 2 ln(10) [(10^x)² - 10^(-x)²]

    dy/dx = 2 ln(10) [10^(2x) - 10^(-2x)]

  • 1 month ago

    (10^x + 10^-x)^2

    = 10^x^2 + 2 + 10^-x^2 

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