Question

Question #91456

Answer 1

`dy/dx = 2x^2sin(x^4) - sinx/(2root(4)(x)) " "" "" "" "x >= 0`

Explanation:

`y = int_sqrtx^(x^2)sqrttsin(t^2)dt = int_0^(x^2)sqrttsin(t^2)dt - int_0^sqrtxsqrttsin(t^2)dt`

We know that for an arbitrary constant `k`:

`d/dx(int_k^xf(t)dt) = f(x)`

Therefore, using chain rule we can also say that

`d/dx(int_k^(g(x))f(t)dt) = f(g(x)) * g'(x)`

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This means that:

`d/dx(int_0^(x^2)sqrttsin(t^2)dt) = sqrt(x^2)sin((x^2)^2)*2x = 2x|x|sin(x^4)`

`d/dx(int_0^sqrtxsqrttsin(t^2)dt) = sqrt(sqrtx)sin(sqrtx^2)*1/(2sqrtx) = sin(x)/(2root(4)(x)`

Note that the second integral is only defined for non-negative values of `x`, since the square root of a negative number is not real.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Therefore, the derivative of `y` is:

`dy/dx = 2x|x|sin(x^4) - sin(x)/(2root(4)(x)`

And since the function is already only defined for non-negative values of `x`, we can replace `absx` with `x`.

`dy/dx = 2x^2sin(x^4) - sinx/(2root(4)(x)) " "" "" "" "x >= 0`

Final Answer