Finding dy/dx of given functions

Question

How do I find dy/dx of the following functions, without simplifying the answers:

a) y = (e^x + (1/x))ln(x²-e^2x)

b) y = ((x^3/5)+(x^1/2))/x

Answer 1

For part (a), we'll have to use a combination of the product rule and the chain rule. Since the quantities e^x + 1/x and ln(x² - e^2x) are being multiplied together, we can use the product rule on them. Recall that the product rule goes like this: (uv)' = uv' + u'v, where (uv)' is the derivative of u times v, and u' and v' are the derivatives of u and v, respectively. In this case: u = e^x + 1/x and v = ln(x² - e^2x).

And for the derivative of ln(x² - e^2x), we'll need to use the chain rule, because there is a function within the ln() function. It goes like this: u(v)' = u'(v) * v', where u is a function that takes v as an input, and v is another function.

u' = (e^x + 1/x)' = e^x - 1/x²

v' = (ln(x² - e^2x))' = 1/(x² - e^2x) * (2x - 2e^2x) = (2x - 2e^2x)/(x² - e^2x)

By combining u, v, u', and v' in the arrangement of the product rule, we can get the final answer:

y' = (e^x + 1/x)(2x - 2e^2x)/(x² - e^2x) + (e^x - 1/x²)ln(x² - e^2x).

Part (b) is much simpler to do. Since only powers of x are involved, we can just use the power rule after a small tweak to the function:

y = x^{3/5 - 1} + x^{1/2 - 1} = x^-2/5 + x^-1/2.

Now by applying the power rule, which says that (x^a)' = ax^a-1 (assuming a is a nonzero real number), we get the final answer:

y' = -2/5 * x^{-2/5 - 1} - 1/2 * x^{-1/2 - 1} = -2x^-7/5/5 - x^-3/2/2.

Without dividing out by x initially, you would have to use the quotient rule. It's not the best way to do this problem, but you would get the same answer if done correctly.

I hope this helps!