HOW TO DERIVATE FUNCTIONS GIVEN IN PARAMETRIC FORM? DERIVATIVES OF TRIGONOMETRIC FUNCTIONS GIVEN IN PARAMETRIC FORM.

To solve this type of derivatives, the chain rule must be used. In these functions there is a relation of the variables x and y through a third variable that plays a fundamental role in this relationship. In this case given a function in the form y = f (x) the parameters x and y depend on a parameter such as t. Let's see an example:

To express functions given in parametric form is done as follows
X =X(t)
Y = Y(t), where t t € IR.  So if f is a derivable function with respect to x and x is a derivable function with respect to t then the variable y can be derived with respect to t by applying the rule of the chain.This is dy / dt = (dy / dx). (Dx / dt), because the variable y depends on x and the variable x depends on t that is y (t) = y (x (t)), then dy / dx = (dy / dt) / (dx / dt) where (dx / dt) ≠ 0 and this expression allows finding the derivative of a given function in parametric form without having to have the explicit representation of said function. Let the trigonometric functions defined in parametric form calculate y' without eliminating the parameter t:

X= a sin(t)   dy/dt = -a sin(t)

Y= a cos(t)   dx/dt = a cos(t) then y'= dy/dx=(-a sen(t))/(a cos(t))

X= t Ln(t)                   dx/dt = Lnt + t(1/t) = Lnt + 1

Y= 2 + cos(t)               dy/dt = –sint  then y'=dy/dx=(-sint)/(Lnt +1)

X= t sin(t)             dx/dt = (sint + t.cost)  

Y= t cos(t)          dy/dt = (cost - t.sent)  then: y'=(cost - t.sent)/(sent + t.cost)

X= 5 + cos(t)     dx/dt = (-sent)  

Y= 6 -  3sin(t)   dy/dt = ( - 3cost)      y'=(-3cost)/(-sent)= 3cotang(t)

X= cotg(t)    dx/dt = (-cosec² (t))  

X= tg(t)        dy/dt = (sec² (t))  y'=( sec² (t))/(-cosec² (t))

Remember that what you must do is calculate the derivatives of "x" and "y" with respect to (t) and then perform the division of the derivative of "y" between the derivative of "x" to express the derivative of "y" "with respect to" x "or what is just same dy / dx.

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