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We include the constant to figure out the derivative in y=6x^2, but we ignore the constant in y=2sin(x). Is it because the constant is not really part of "x" in the second equation?

2007-08-17 19:53:12 · 5 answers · asked by Muffins 1 in Science & Mathematics Mathematics

5 answers

y = 6 x²
dy/dx = 6 (2x) = 12 x

y = 2 sin x
dy / dx = 2 cos x

In the first question it is x² that is differentiated and in the second question it is cos x that is differentiated.

Rules for above are:-
y = k x²
dy/dx = k (d/dx) x² = 2 k x

y = k sin x
dy/dx = k (d/dx)(sin x) = k cos x

2007-08-18 04:47:56 · answer #1 · answered by Como 7 · 1 0

Basically derivative means the rate of change of something with respect to another particular thing.

so, in case of constant term , as it is constant the rate of change with respect to anything would always be zero.
Thus we can ignore that constant term in each and every cases and actually we always follow this rule. But if that constant term remains in the problem as the multiplying factor of a variable term then we can't ignore that term. In that case 1st we have to find out the derivative of that variable term and then we have to multiply that value with that constant term.

so, in ur 1st problem as the constant term i.e. 6 is a multiplying factor of the variable term i.e. x^2 , we have to consider that constant term and thus the answer would be 12x. In this case if we neglect that constant term then we can't get the proper answer. The value of gradient of the corresponding curve of the equation will be changed. Because we can say that the derivative of any equation is nothing but the gradient of that corresponding curve.

In the 2nd problem u can use the same logic to solve the derivative. I think this will help u much to solve any derivatives.

2007-08-18 03:32:23 · answer #2 · answered by sharbadeb 2 · 0 0

In actuality, the constant is still there, but just merged with the result of the derivative. In essence, you're talking about the property

d/dx c f(x) = c f'(x)

Using your examples:

1) f(x) = 6x^2

f'(x) = 6 (2x)
But, we can multiply out the 6 and the 2, to get
f'(x) = 12x

2) y = 2sin(x)
dy/dx = 2 cos(x)

2007-08-18 02:58:08 · answer #3 · answered by Puggy 7 · 3 1

The two "2"s have different meanings.
x^2 means x*x
dx^2 /dx = dx*x/dx = x + x = 2x, "2" here changes from the exponential to a constant factor.

2sin(x) means sin(x) + sin(x)
d2sin(x) dx = cos(x) + cos(x) = 2cos(x), "2" here is still a constant factor

2007-08-18 02:58:37 · answer #4 · answered by sahsjing 7 · 1 0

guy be clear differentiation is an operation to be applied only on functions and in the first the case is completely different.do the problem by first principles say y= f(x) , y+deltay =f(x+deltax) now del y=f(x+ del x)-f(x)you will get the del y in first eqn as 12*delx *x + delx^2.you can neglect second term and apply limit and that solves the problem.

2007-08-18 03:56:09 · answer #5 · answered by astro_crabnebulae 1 · 0 0

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