Exercise
$\frac{d}{dx}cot^2\left(x-y\right)+\sec^2y=\:tan^2x$
Step-by-step Solution
Learn how to solve problems step by step online. Find the implicit derivative d/dx(cot(x-y)^2+sec(y)^2=tan(x)^2). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1}. Any expression to the power of 1 is equal to that same expression. 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)}.
Find the implicit derivative d/dx(cot(x-y)^2+sec(y)^2=tan(x)^2)
Final answer to the exercise
$y^{\prime}=\frac{\tan\left(x\right)\sec\left(x\right)^2+\cot\left(x-y\right)\csc\left(x-y\right)^2}{\cot\left(x-y\right)\csc\left(x-y\right)^2+\sec\left(y\right)^2\tan\left(y\right)}$