Find the derivative of the function. y = sin(x) ln(8 + 3v) dv cos(x)

Expert Answers

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First, it depends on how you interpret the `dv` term. The most obvious way is to interpret it as `(dv)/dx` which is how I'll treat it. There are two things to remember with regard to differential calculus when looking at this question: the product rule and the chain rule:

`d/(dx) (UV) = (dU)/dx * V + (dV)/dx * U` - Product rule

`d/dx F(g(x)) = (dF)/(dg) * (dg)/(dx)` - Chain rule

The product rule comes into play when we see the above as a set of 4 functions of x multiplied together:

`f_1(x) = sin(x)`

`f_2(x) = ln(8+3v)`

`f_3(x) = (dv)/(dx)`

`f_4(x) = cos(x)`

Which means the following about their derivatives:

`d/dx f_1 = cos(x)`

`d/dxf_2 = 1/(8+3v) * 3 * (dv)/(dx)`


`d/dx(f_4) = -sin(x)`

Note, for `f_2` you must first interpret it as `ln(u(v))` where `u(v) = 8+3v` , so you use the chain rule to get the following:

`(df_2)/(dx) = d/(du) ln(u) * (du)/(dv) * (dv)/(dx) = 1/u * 3 * (dv)/(dx)`

This result simplifies to our above calculated derivative.

The product rule for the original problem can be generalized in the following way:

`(dy)/dx = (df_1)/dx * f_2 * f_3 * f_4 + f_1 * (df_2)/dx * f_3 * f_4 + f_1*f_2*(df_3)/dx *f_4 + f_1 * f_2 * f_3 * (df_4)/dx`

Now, we just plug in values for each of the 4 terms:

`(df_1)/dx * f_2 * f_3 * f_4 = cos(x) * ln(8+3v) * dv * cos(x)`

`f_1 * (df_2)/(dx) * f_3 * f_4 = sin(x) * 3/(8+3v) * (dv)/(dx) * (dv)/(dx) * cos(x) `

`f_1 * f_2 * (df_3)/(dx) * f_4 = sin(x) * ln(8+3v) * (d^2v)/(dx^2)*cos(x)`

`f_1 * f_2 * f_3 * (df_4)/dx = sin(x)*ln(8+3v) * dv * -sin(x)`

You can now simplify the sum with trigonometric identities (recognize the first term and last term can be simplified to a useful form by combining their sum and removing the common terms outside of the parentheses):

`ln(8+3v)dv * (cos^2(x) - sin^2(x))= ln(8+3v)dv cos(2x)`

Similarly the middle two terms can be put into terms with respect to `sin(2x)` , but it does not allow for simplification of the overall equation, so there's not much of a reason to do it. Your final result is the following when taking all of the summed terms:

`(dy)/dx = ln(8+3v)(dv)/dx*cos(2x) + sin(x)cos(x) * (3/(8+3v) * ((dv)/(dx))^2 + ln(8+3v)*(d^2v)/(dx^2))`

It might make for a cleaner solution if `v` is a constant with respect to `x` , but then you get a trivial result because `dv=0` .

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