
How do you find the derivative of $f(t) = 4t$?
Answer
533.1k+ views
Hint: In this question, we need to differentiate the given function with respect to the variable $t$. Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable. Note that for the given function, we can use the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$. So find the value of $n$ and simplify using the formula to obtain the derivative of the given function.
Complete step-by-step solution:
Given a function of the form $f(t) = 4t$
We are asked to find out the derivative of the above function.
i.e. we need to differentiate the given function with respect to the variable $t$.
Firstly, let us understand the meaning of differentiation.
Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable per unit modification.
Let $y = f(x)$ be a function of x. Then the rate of change of y per unit change in x is given by, $\dfrac{{dy}}{{dx}}$.
Note that the differentiation is the method of evaluating a function’s derivative at any time.
Now let us consider the given function $f(t) = 4t$.
We know that the derivative of ${x^n}$ with respect to x is $n{x^{n - 1}}$.
i.e. $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
In the given problem, we have $n = 1$.
Hence differentiating the given function with respect to $t$, we get,
$ \Rightarrow \dfrac{d}{{dt}}(f(t)) = \dfrac{d}{{dt}}(4t)$
Simplifying this we use the formula, $\dfrac{d}{{dx}}cu = c\dfrac{d}{{dx}}u$ where c is a constant.
Here $c = 4$. So taking out the constant 4 we get,
$ \Rightarrow 4\dfrac{d}{{dt}}{(t)^1}$
Now using the formula we have,
$ \Rightarrow 4{(t)^{1 - 1}}$
$ \Rightarrow 4{t^0}$
We know that ${t^0} = 1$
Hence we get,
$ \Rightarrow 4 \times 1 = 4$
Thus, the derivative of the function $f(t) = 4t$ is $4$.
Note: The process to find the derivative of the function is called differentiation. In differentiation, there is an instantaneous rate of change of the functions based on the variable.
Students must remember some formulas of differentiation to find the derivative of a function.
Some of the basic differentiation formulas are as follows.
(1) $\dfrac{d}{{dx}}c = 0$ where, c is a constant.
(2) $\dfrac{d}{{dx}}x = 1$
(3) $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
(4) $\dfrac{d}{{dx}}cu = c\dfrac{d}{{dx}}u$ where, c is a constant.
(5) $\dfrac{d}{{dx}}u \pm v = \dfrac{d}{{dx}}u \pm \dfrac{d}{{dx}}v$
Complete step-by-step solution:
Given a function of the form $f(t) = 4t$
We are asked to find out the derivative of the above function.
i.e. we need to differentiate the given function with respect to the variable $t$.
Firstly, let us understand the meaning of differentiation.
Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable per unit modification.
Let $y = f(x)$ be a function of x. Then the rate of change of y per unit change in x is given by, $\dfrac{{dy}}{{dx}}$.
Note that the differentiation is the method of evaluating a function’s derivative at any time.
Now let us consider the given function $f(t) = 4t$.
We know that the derivative of ${x^n}$ with respect to x is $n{x^{n - 1}}$.
i.e. $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
In the given problem, we have $n = 1$.
Hence differentiating the given function with respect to $t$, we get,
$ \Rightarrow \dfrac{d}{{dt}}(f(t)) = \dfrac{d}{{dt}}(4t)$
Simplifying this we use the formula, $\dfrac{d}{{dx}}cu = c\dfrac{d}{{dx}}u$ where c is a constant.
Here $c = 4$. So taking out the constant 4 we get,
$ \Rightarrow 4\dfrac{d}{{dt}}{(t)^1}$
Now using the formula we have,
$ \Rightarrow 4{(t)^{1 - 1}}$
$ \Rightarrow 4{t^0}$
We know that ${t^0} = 1$
Hence we get,
$ \Rightarrow 4 \times 1 = 4$
Thus, the derivative of the function $f(t) = 4t$ is $4$.
Note: The process to find the derivative of the function is called differentiation. In differentiation, there is an instantaneous rate of change of the functions based on the variable.
Students must remember some formulas of differentiation to find the derivative of a function.
Some of the basic differentiation formulas are as follows.
(1) $\dfrac{d}{{dx}}c = 0$ where, c is a constant.
(2) $\dfrac{d}{{dx}}x = 1$
(3) $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
(4) $\dfrac{d}{{dx}}cu = c\dfrac{d}{{dx}}u$ where, c is a constant.
(5) $\dfrac{d}{{dx}}u \pm v = \dfrac{d}{{dx}}u \pm \dfrac{d}{{dx}}v$
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