Derivatives Calculus 1

Question

I just need help with this derivative. X(t)= (1+t)/sin(t)

Please show all steps! Thank you!

Answer 1

d/dt (1+ t) / sin(t) = d/(dt) (1 + t) * csc(t).

From here we can use the product rule:

d/dt uv = v * du/dt + u * dv/dt, where v = csc(t), u = t + 1.

So, now our expression looks like this:

csc(t) * (d/dt (1 + t)) + (1 + t) * (d/dt csc(t)).

From here we differentiate term-by-term:

csc(t) * 1 + (1 + t) * (d/dt csc(t)) =

csc(t) + (1 + t) * (d/dt csc(t)) =

Now, using the chain rule, we turn d/dt csc(t) into (d csc(x) / dx) * (dt/dx), where x = t and (d/dx csc(x)) = - cot(x)csc(x). So,

csc(t) + (1 + t) * -cot(t)csc(t) * d/dt t =

csc(t) + (1 + t) * -cot(t)csc(t) * 1 =

csc(t) + (1 + t) * -cot(t)csc(t) =

csc(t) - (1 + t) * cot(t)csc(t)

Answer 2

Hi Brett,

it is enough you follow the rule of derivative of fraction

f(x) = u/v f'(x) = ( u' v - u v') / v²

Now, you have X(t)= (1+t)/sin(t) it means: u = 1 + t and v = sin(t)

u' = 1 and v'= cos(t)

now replace in the formula and be careful that function is x and variable is t

X'(t) = [1 * sin(t) - (1+t)cos(t)] / sin²(t)

X'(t) = [ sin(t) - cos(t) - t cos(t)] / sin²(t)

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My friend, when you need to find derivative of any function

1) recognize the kind of function( such as fraction, radical, ...)

2) follow the rule of that function, if it is complicate, in separately find the derivative then use the main formula.

I hope it is useful,

Minoo