Question

How do you find the derivative of $P\left( t \right)=3000+500\sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)$?

Answer 1

Consider the argument of the sine function as f (t) and write the given function P (t) in the form: $P\left( t \right)=3000+500\sin \left( f\left( t \right) \right)$. Now, differentiate both the sides with respect to the variable t and use the chain rule of differentiation given as: $\dfrac{d\left[ \sin \left( f\left( t \right) \right) \right]}{dt}=\cos \left( f\left( t \right) \right)\times f'\left( t \right)$ to find the derivative of P (t). Here, f’(t) is the derivative of the assumed function f (t). Use the fact that the derivative of a constant term is 0.

Complete step by step solution:
Here, we have been provided with the function $P\left( t \right)=3000+500\sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)$ and we are asked to differentiate it. Here we are going to use the chain rule of differentiation to get the answer.
$\because P\left( t \right)=3000+500\sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)$
Clearly, we can see that we have P (t) as a function of variable t. Now, we have another function as the argument of the sine function, so let us assume the argument of sine as f (t). So, we have,
$P\left( t \right)=3000+500\sin \left( f\left( t \right) \right)$, here $f\left( t \right)=\left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)$.
Now, we can assume the above function as a composite function, so we need to use the chain rule of differentiation to get the required derivative. Therefore, using the formula:
$\dfrac{d\left[ \sin \left( f\left( t \right) \right) \right]}{dt}=\dfrac{d\left[ \sin \left( f\left( t \right) \right) \right]}{d\left( f\left( t \right) \right)}\times \dfrac{d\left( f\left( t \right) \right)}{dt}$, we have on differentiating both sides with respect to t,
$\Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=\dfrac{d\left( 3000 \right)}{dt}+\dfrac{d\left( 500\left( \sin \left( f\left( t \right) \right) \right) \right)}{d\left( f\left( t \right) \right)}\times \dfrac{d\left( f\left( t \right) \right)}{dt}$
Substituting back the assumed value of f (t), we get,
$\Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=\dfrac{d\left( 3000 \right)}{dt}+\dfrac{d\left( 500\left( \sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right) \right)}{d\left[ \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right]}\times \dfrac{d\left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)}{dt}$
Now, using the fact that the derivative of a constant term is 0, we get,
$\Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=\dfrac{d\left( 500\left( \sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right) \right)}{d\left[ \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right]}\times \dfrac{d\left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)}{dt}$
Since 500 is a constant multiplied to the function, so it can be taken out of the derivative. Using the formula: $\dfrac{d\left[ \sin \left( f\left( t \right) \right) \right]}{dt}=\cos \left( f\left( t \right) \right)\times f'\left( t \right)$, we get,

& \Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=500\times \dfrac{d\left( \sin \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right)}{d\left[ \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \right]}\times \dfrac{d\left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)}{dt} \\\ & \Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=500\times \cos \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)\times \left[ \dfrac{d\left( 2\pi t \right)}{dt}-\dfrac{d\left( \dfrac{\pi }{2} \right)}{dt} \right] \\\ & \Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=500\times \cos \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right)\times \left[ 2\pi \right] \\\ & \Rightarrow \dfrac{d\left( P\left( t \right) \right)}{dt}=1000\pi \cos \left( 2\pi t-\left( \dfrac{\pi }{2} \right) \right) \\\ \end{aligned}$$ Hence, the above relation is our answer. **Note:** Remember the basic formulas of derivatives of all the trigonometric functions because they are used in integral calculus also. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, $$\dfrac{u}{v}$$ rule etc. as they are frequently used in both differential and integral calculus. Note that whenever we have a composite function we use the chain rule to find its derivative.