Home > Calculus > Derivative of Parametric Functions

Find the derivative of #1/(-sint)#?

I've taken the negative sign out to get #-(1/\sint)'# but that's as far as I got...

I wanted to apply the product rule instead of the quotient rule.
(Symbolab uses the quotient rule)

Answer 1

Emily Ammerman

#cot(t)csc(t)#

Before starting we know that #-1/sint=-csct#

Also

#-d/dtcsct=-(-cot(t)csc(t))#

#=cot(t)csc(t)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

To find the derivative of ( \frac{1}{-\sin(t)} ), we can use the chain rule and the derivative of the sine function.

[ \frac{d}{dt} \left( \frac{1}{-\sin(t)} \right) = -\frac{d}{dt} \left( \frac{1}{\sin(t)} \right) ]

Using the chain rule, we have:

[ -\frac{d}{dt} \left( \frac{1}{\sin(t)} \right) = -\frac{d}{d(\sin(t))} \left( \frac{1}{\sin(t)} \right) \frac{d(\sin(t))}{dt} ]

[ = -\left( -\frac{1}{\sin^2(t)} \right) \cos(t) ]

[ = \frac{\cos(t)}{\sin^2(t)} ]

So, the derivative of ( \frac{1}{-\sin(t)} ) with respect to ( t ) is ( \frac{\cos(t)}{\sin^2(t)} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Find the derivative of #1/(-sint)#?

I've taken the negative sign out to get #-(1/\sint)'# but that's as far as I got... I wanted to apply the product rule instead of the quotient rule. (Symbolab uses the quotient rule)

I've taken the negative sign out to get #-(1/\sint)'# but that's as far as I got...

I wanted to apply the product rule instead of the quotient rule.
(Symbolab uses the quotient rule)