Question

How do you differentiate `y= sqrt(x) e^(x^2) (x^2+3)^5`?

Answer 1

`dy/dx = (sqrt(x)e^(x^2)(x^2 + 3)^5)(1/(2x) + 2x + (10x)/(x^2 + 3))`

Explanation:

I would use logarithmic differentiation.

`lny = ln(sqrt(x)e^(x^2)(x^2 + 3)^5))`

Using `ln(ab) = ln(a) + ln(b)`.

`lny = lnsqrt(x) + ln(e^(x^2)) + ln(x^2 + 3)^5`

`lny = lnx^(1/2) + ln(e^(x^2)) + ln(x^2 + 3)^5`

Now we use `lna^n = nlna`.

`lny = 1/2lnx + x^2ln(e) + 5ln(x^2 + 3)`

`lny = 1/2lnx + x^2 + 5ln(x^2 +3)`

Now the derivative is given by the chain rules and `d/dx(lnx) = 1/x`.

`1/y(dy/dx) = 1/(2x) + 2x + (5(2x))/(x^2 + 3)`

`dy/dx= y(1/(2x) + 2x + (10x)/(x^2 + 3))`

`dy/dx = (sqrt(x)e^(x^2)(x^2 + 3)^5)(1/(2x) + 2x + (10x)/(x^2 + 3))`

Hopefully this helps!