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Use logarithmic differentiation to differentiate the function with respect to x.  y=(x^5+5)^2  sqrt(2x^2+3) ?

I got the answer 

y'=(2x(x^5+5) (11x^5+15x^3+5))/sqrt(2x^2+3) 

But I need to know the steps to get there. 

Update:

UPDATE: I did not thumbs down llafer's answer, I made it the favorite and thumbs up'd it. 

2 Answers

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  • 1 month ago
    Favorite Answer

    I got the steps from the page linked below.  But I'll explain them here as well.

    Starting with your equation:

    y = (x⁵ + 5)² √(2x² + 3)

    Get the log of both sides:

    ln(y) = ln[(x⁵ + 5)² √(2x² + 3)]

    Now break up that log into the sum/differences of other logs.  Starting out with the log of a product is the same as the sum of two logs:

    ln(y) = ln[(x⁵ + 5)²] + ln[√(2x² + 3)]

    Now we do an implicit differentiation of both sides with respect to x.

    Recall that the derivative of y = ln(x) is y' = 1/x

    The left side becomes:

    y' / y

    The two terms on the left side each will need to use the chain rules at least once.  Let's do them separately then piece them back together:

    term 1:

    y = ln[(x⁵ + 5)²] can be written as y = ln(u) and u = (x⁵ + 5)². We'll need the chain rule again:

    u = v² and v = x⁵ + 5

    Derivative of both of these:

    du/dv = 2v and dv/dx = 5x⁴

    chain rule:

    du/dx = du/dv * dv/dx

    du/dx = 2v * 5x⁴

    du/dx = 10vx⁴

    du/dx = 10(x⁵ + 5)x⁴

    Now that we have that we can do the first chain rule:

    y = ln(u) and u = (x⁵ + 5)²

    dy/du = 1/u and du/dx = 10(x⁵ + 5)x⁴

    dy/dx = dy/du * du/dx

    dy/dx = 1/u * 10(x⁵ + 5)x⁴

    dy/dx = 10(x⁵ + 5)x⁴ / u

    dy/dx = 10(x⁵ + 5)x⁴ / (x⁵ + 5)²

    The common factor can cancel out leaving:

    dy/dx = 10x⁴ / (x⁵ + 5)

    term 2:

    Same steps as term 1. We'll need two chan rules:

    y = ln[√(2x² + 3)]

    y = ln(u) and u = √(2x² + 3)

    u = √v and v = 2x² + 3

    du/dv = 1/(2√v) and dv/dx = 4x

    du/dx = du/dv * dv/dx

    du/dx = 1 / (2√v) * 4x

    du/dx = 4x / (2√v)

    du/dx = 2x / √v

    du/dx = 2x / √(2x² + 3)

    Back to the first chain:

    y = ln(u) and u = √(2x² + 3)

    dy/du = 1/u and du/dx = 2x / √(2x² + 3)

    dy/dx = dy/du * du/dx

    dy/dx = 1/u * 2x / √(2x² + 3)

    dy/dx = 2x / [u√(2x² + 3)]

    dy/dx = 2x / [√(2x² + 3) √(2x² + 3)]

    We have the square root of a square in the denominator so this simplifies:

    dy/dx = 2x / (2x² + 3)

    Now we can build our equation again after the implicit differentiation step on both sides of the equation:

    y'/y = 10x⁴ / (x⁵ + 5) + 2x / (2x² + 3)

    Now we multiply both sides by y:

    y' = y[10x⁴ / (x⁵ + 5) + 2x / (2x² + 3)]

    And then substitute the initial value for y from the start of the problem and simplify:

    y' = (x⁵ + 5)² √(2x² + 3) * [10x⁴ / (x⁵ + 5) + 2x / (2x² + 3)]

    If we distribute that "y" expression into both terms:

    y' = (x⁵ + 5)² √(2x² + 3) (10x⁴) / (x⁵ + 5) + (x⁵ + 5)² √(2x² + 3)(2x) / (2x² + 3)

    the (x⁵ + 5)'s in the first term cancel out along with √(2x² + 3) in the second term resulting in a square root in the denominator:

    y' = (x⁵ + 5) √(2x² + 3) (10x⁴) + (x⁵ + 5)² (2x) / √(2x² + 3)

    If we get a common denominator now, the square root in the first term becomes rational:

    y' = (x⁵ + 5)(2x² + 3)(10x⁴) / √(2x² + 3) + (x⁵ + 5)² (2x) / √(2x² + 3)

    And now we can add the numerators:

    y' = [(x⁵ + 5)(2x² + 3)(10x⁴) + (x⁵ + 5)² (2x)] / √(2x² + 3)

    We can factor a 2x and (x⁵ + 5) from the numerator:

    y' = 2x(x⁵ + 5)[(2x² + 3)(5x³) + (x⁵ + 5)] / √(2x² + 3)

    And we'll simplify what's left in that third factor:

    y' = 2x(x⁵ + 5)(10x⁵ + 15x³ + x⁵ + 5) / √(2x² + 3)

    y' = 2x(x⁵ + 5)(11x⁵ + 15x³ + 5) / √(2x² + 3)

    And that's the answer that you were able to look up and all the steps to get there.

  • 1 month ago

    The answer given by llaffer is top class, so begs the question why a thumbs down? Anyway, a TU from me as it cannot be bettered.

    :)>

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