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Question: How do you find the derivative of \(f(t) = - 2{t^2} + 3t - 6\)?...

How do you find the derivative of f(t)=−2t2+3t−6f(t) = - 2{t^2} + 3t - 6?

Explanation

Solution

In this question, we need to differentiate the given function with respect to the variable t. Differentiation can be defined as a derivative of an independent variable value and can be used to calculate features in an independent variable. Note that for the given function, we can use the power rule of differentiation d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}. So find the value of n and simplify using the formula to obtain the derivative of the given function.

Complete step by step answer:
We have to evaluate the derivative of f(t)=−2t2+3t−6f(t) = - 2{t^2} + 3t - 6 using the power rule of differentiation.
So, to evaluate the derivative, we have to differentiate the function with respect to t.
We will be using the power rule of differentiation to evaluate the derivative of the given function.
So, we have, f(t)=−2t2+3t−6f(t) = - 2{t^2} + 3t - 6.
Differentiating both sides with respect to the variable t, we get,
⇒f′(t)=ddt(−2t2+3t−6)\Rightarrow f'(t) = \dfrac{d}{{dt}}\left( { - 2{t^2} + 3t - 6} \right)
Now, separating the differentiation of all the terms, we get,
⇒f′(t)=ddt(−2t2)+ddt(3t)−ddt(6)\Rightarrow f'(t) = \dfrac{d}{{dt}}\left( { - 2{t^2}} \right) + \dfrac{d}{{dt}}\left( {3t} \right) - \dfrac{d}{{dt}}\left( 6 \right)
Now, we know that the derivative of the constant term with respect to a variable is zero. Hence, we get,
⇒f′(t)=ddt(−2t2)+ddt(3t)−0\Rightarrow f'(t) = \dfrac{d}{{dt}}\left( { - 2{t^2}} \right) + \dfrac{d}{{dt}}\left( {3t} \right) - 0
Now, we take the constants out of the differentiation, we get,
⇒f′(t)=−2ddt(t2)+3ddt(t)\Rightarrow f'(t) = - 2\dfrac{d}{{dt}}\left( {{t^2}} \right) + 3\dfrac{d}{{dt}}\left( t \right)
Now, using the power rule of differentiation d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}, we get,
⇒f′(t)=−2(2t)+3(1)\Rightarrow f'(t) = - 2\left( {2t} \right) + 3\left( 1 \right)
Simplifying the expression, we get,
⇒f′(t)=−4t+3\Rightarrow f'(t) = - 4t + 3

Note:
The derivative of a constant is always equal to zero and we can take out the coefficients of the terms outside of the differentiation as d(k×f(x))dx=k×d(f(x))dx\dfrac{{d\left( {k \times f\left( x \right)} \right)}}{{dx}} = k \times \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}. Power rule of differentiation helps us to differentiate such power functions xn{x^n} with respect to x as nxn−1n{x^{n - 1}}.