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Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
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$\frac{\frac{d}{dx}\left(\tan\left(x\right)\right)\csc\left(2x^3\right)-\tan\left(x\right)\frac{d}{dx}\left(\csc\left(2x^3\right)\right)}{\csc\left(2x^3\right)^2}$
Learn how to solve quotient rule of differentiation problems step by step online. Find the derivative d/dx(tan(x)/csc(2x^3)). Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if {f(x) = tan(x)}, then {f'(x) = sec^2(x)\cdot D_x(x)}. The derivative of the linear function is equal to 1. Taking the derivative of cosecant function: \frac{d}{dx}\left(\csc(x)\right)=-\csc(x)\cdot\cot(x)\cdot D_x(x).