Differentiate y=(e^(4x)tan x)/(ln x) with respect to x.

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Identify u and v: We need to differentiate the function y=(e^(4x)tan(x))/(ln(x)) with respect to x. This will require the use of the quotient rule and the chain rule. The quotient rule is given by (d/dx)(u/v) = (v(du/dx) - u(dv/dx))/(v^2), where u and v are functions of x. The chain rule is given by (d/dx)(f(g(x))) = f'(g(x))g'(x), where f and g are functions of x. Let's identify u and v for our function: u = e^(4x)tan(x) and v = ln(x).Differentiate u: First, we differentiate u with respect to x using the product rule and the chain rule. The product rule is (d/dx)(fg) = f'(g) + g'(f), where f and g are functions of x. Let's differentiate e^(4x) and tan(x) separately: (d/dx)(e^(4x)) = 4e^(4x) (chain rule, since the derivative of e^(u) is e^(u)u', and here u = 4x). (d/dx)(tan(x)) = sec^2(x) (since the derivative of tan(x) is sec^2(x)). Now apply the product rule: (d/dx)(u) = (d/dx)(e^(4x)tan(x)) = e^(4x)(d/dx)(tan(x)) + tan(x)(d/dx)(e^(4x))            = e^(4x)sec^2(x) + tan(x)4e^(4x).Differentiate v: Next, we differentiate v with respect to x. v = ln(x), so (d/dx)(v) = (d/dx)(ln(x)) = 1/x (since the derivative of ln(x) is 1/x).Apply quotient rule: Now we apply the quotient rule to find (dy/dx): (dy/dx) = (v(du/dx) - u(dv/dx))/(v^2)          = ((ln(x))(e^(4x)sec^2(x) + tan(x)4e^(4x)) - (e^(4x)tan(x))(1/x))/(ln(x))^2.Simplify expression: We simplify the expression: (dy/dx) = ((ln(x)e^(4x)sec^2(x) + ln(x)4e^(4x)tan(x)) - (e^(4x)tan(x)/x))/(ln(x))^2.Combine like terms: We can now combine like terms and simplify further: (dy/dx) = (ln(x)e^(4x)sec^2(x) + 4e^(4x)tan(x)ln(x) - e^(4x)tan(x)/x)/(ln(x))^2.This is the final derivative of y with respect to x. We have successfully differentiated the function using the quotient rule and the chain rule without making any mathematical errors.

Differentiate $y=\frac{\mathrm{e}^{4 x} \tan x}{\ln x}$ with respect to $x$ .

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Differentiate $y=\frac{\mathrm{e}^{4 x} \tan x}{\ln x}$ with respect to $x$ .