Asked by Ann
                When you do implicit differentiation, how does D = √(x^2 + 8x + 12) turn into dD/dt = [(x + 4)(dx/dt)]/√(x^2 + 8x + 12)?
please explain...I don't even understand where the (x+4) comes from. D means distance, but that's irrelevant
            
        please explain...I don't even understand where the (x+4) comes from. D means distance, but that's irrelevant
Answers
                    Answered by
            Steve
            
    It's really just the chain rule in disguise:
D = √u
where u is a function of x
dD/dx = 1/(2√u) du/dx
du/dx = 2x+8
Now, since x is a function of t,
dD/dt = dD/dx dx/dt
and voila
    
D = √u
where u is a function of x
dD/dx = 1/(2√u) du/dx
du/dx = 2x+8
Now, since x is a function of t,
dD/dt = dD/dx dx/dt
and voila
                    Answered by
            Jai
            
    D = √(x^2 + 8x + 12)
We can also rewrite this as
D = (x^2 + 8x + 12)^(1/2)
Now we have to get dD/dt. To get the derivative, recall that if x is raised to a certain constant, the derivative is
x^n = n*x^(n-1) * dx
We multiply the exponent to x, which is jow raised to 1 less than the original exponent, and multiplied by the derivative of x.
For example,
derivative of (2x)^5 = 5*(3x)^4 * 3 = 15*(3x)^4
Don't forget that you have to multiply the derivative of 3x (which is 3) to the whole expression.
Therefore, in the problem, dD/dt((x^2 + 8x + 12)^(1/2))
= ((1/2)*(x^2 + 8x + 12)^(1/2 - 1) * d/dx((x^2 + 8x + 12)))*dx/dt
We know that the derivative of the x^2 + 8x + 12 is 2x + 8:
= (1/2)*(2x + 8)*(x^2 + 8x + 12)^(-1/2) dx/dt
= (x + 4)*(x^2 + 8x + 12)^(-1/2) dx/dt
Note that a term raised by negative exponent can be rewritten as a term raised with positive exponent, but now placed in the denominator. Rewriting this expression, you'll get the answer you typed there:
= [(x + 4)/√(x^2 + 8x + 12)](dx/dt)
Hope this helps :3
    
We can also rewrite this as
D = (x^2 + 8x + 12)^(1/2)
Now we have to get dD/dt. To get the derivative, recall that if x is raised to a certain constant, the derivative is
x^n = n*x^(n-1) * dx
We multiply the exponent to x, which is jow raised to 1 less than the original exponent, and multiplied by the derivative of x.
For example,
derivative of (2x)^5 = 5*(3x)^4 * 3 = 15*(3x)^4
Don't forget that you have to multiply the derivative of 3x (which is 3) to the whole expression.
Therefore, in the problem, dD/dt((x^2 + 8x + 12)^(1/2))
= ((1/2)*(x^2 + 8x + 12)^(1/2 - 1) * d/dx((x^2 + 8x + 12)))*dx/dt
We know that the derivative of the x^2 + 8x + 12 is 2x + 8:
= (1/2)*(2x + 8)*(x^2 + 8x + 12)^(-1/2) dx/dt
= (x + 4)*(x^2 + 8x + 12)^(-1/2) dx/dt
Note that a term raised by negative exponent can be rewritten as a term raised with positive exponent, but now placed in the denominator. Rewriting this expression, you'll get the answer you typed there:
= [(x + 4)/√(x^2 + 8x + 12)](dx/dt)
Hope this helps :3
                    Answered by
            Reiny
            
    D^2 = (x^2 + 8x + 12)
2D dD/dt = 2x dx/dt + 8 dx/dt
dD/dt = dx/dt (2x+8)/2D
dD/dt = dx/dt (x+4) / √(x^2 + 8x + 12)
    
2D dD/dt = 2x dx/dt + 8 dx/dt
dD/dt = dx/dt (2x+8)/2D
dD/dt = dx/dt (x+4) / √(x^2 + 8x + 12)
                    Answered by
            Steve
            
    Nice one, Reiny. Went right to the heart of the matter of implicit derivatives.
    
                    Answered by
            Ann
            
    Thank you all so much!
    
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