How do you use the definition of a derivative to find the derivative of f(x) = – 3x?
Answer
608.4k+ views
Hint: We are given f(x) = – 3x and we are asked to find the derivative of f(x). To do the same, we will learn about the product rule. We will split our function f(x) = – 3x into two fractions and then apply the product rule where it is given as \[{{\left( uv \right)}^{'}}={{u}^{'}}v+{{v}^{'}}u\] and then we also need \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] to simplify our solution.
Complete step-by-step solution:
We are given a function f(x) = – 3x and we have to differentiate it. Now, we will first observe our function. We can see that f(x) = – 3x is given as the product of – 3 and x. So, we can see that our function is the product of 2 functions in which one is – 3 and the other is x. As we know to find the derivative of the product of 2 functions, we will need the product rule. The product rule is given as \[{{\left( uv \right)}^{'}}={{u}^{'}}v+{{v}^{'}}u.\] Now, in f(x) = – 3x, we consider u = – 3 and v = x. So, applying the product rule on f(x) = – 3x, we get,
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( -3x \right)}{dx}=x\dfrac{d\left( -3 \right)}{dx}+\left( -3 \right)\dfrac{d\left( x \right)}{dx}\]
Now, as we know the derivative of constant is zero, so \[\dfrac{d\left( -3 \right)}{dx}=0\] and we have \[\dfrac{dx}{dx}=1,\] as we compare \[\dfrac{d\left( x \right)}{dx}\] with \[\dfrac{d\left( {{x}^{n}} \right)}{dx}\] then we get n = 1 as \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] and \[\dfrac{dx}{dx}=1{{x}^{1-1}}=1.\]
Now using \[\dfrac{d\left( -3 \right)}{dx}=0\] and \[\dfrac{dx}{dx}=1,\] we get,
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( -3x \right)}{dx}=x\left( 0 \right)+\left( -3 \right)\left( 1 \right)\]
On simplifying, we get,
\[\Rightarrow \dfrac{d\left( -3x \right)}{dx}=-3\]
Here, the derivative of – 3x is – 3.
Note: As we can see that our function is the product of 2 functions of which one is constant, so there is another way to find the derivative. Our function is of the type f(x) = k.g(x), that is constant multiplied by other function and then in the derivative \[\dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( kg\left( x \right) \right)}{dx},\] we can take out k and we get \[d\left( kg\left( x \right) \right)=kd\left( g\left( x \right) \right).\] So, in our case f(x) = – 3x.
\[\dfrac{d\left( -3x \right)}{dx}=-3\dfrac{dx}{dx}=-3\]
As \[\dfrac{dx}{dx}=1.\]
So, the derivative of f(x) = – 3x is – 3.
Complete step-by-step solution:
We are given a function f(x) = – 3x and we have to differentiate it. Now, we will first observe our function. We can see that f(x) = – 3x is given as the product of – 3 and x. So, we can see that our function is the product of 2 functions in which one is – 3 and the other is x. As we know to find the derivative of the product of 2 functions, we will need the product rule. The product rule is given as \[{{\left( uv \right)}^{'}}={{u}^{'}}v+{{v}^{'}}u.\] Now, in f(x) = – 3x, we consider u = – 3 and v = x. So, applying the product rule on f(x) = – 3x, we get,
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( -3x \right)}{dx}=x\dfrac{d\left( -3 \right)}{dx}+\left( -3 \right)\dfrac{d\left( x \right)}{dx}\]
Now, as we know the derivative of constant is zero, so \[\dfrac{d\left( -3 \right)}{dx}=0\] and we have \[\dfrac{dx}{dx}=1,\] as we compare \[\dfrac{d\left( x \right)}{dx}\] with \[\dfrac{d\left( {{x}^{n}} \right)}{dx}\] then we get n = 1 as \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] and \[\dfrac{dx}{dx}=1{{x}^{1-1}}=1.\]
Now using \[\dfrac{d\left( -3 \right)}{dx}=0\] and \[\dfrac{dx}{dx}=1,\] we get,
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( -3x \right)}{dx}=x\left( 0 \right)+\left( -3 \right)\left( 1 \right)\]
On simplifying, we get,
\[\Rightarrow \dfrac{d\left( -3x \right)}{dx}=-3\]
Here, the derivative of – 3x is – 3.
Note: As we can see that our function is the product of 2 functions of which one is constant, so there is another way to find the derivative. Our function is of the type f(x) = k.g(x), that is constant multiplied by other function and then in the derivative \[\dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( kg\left( x \right) \right)}{dx},\] we can take out k and we get \[d\left( kg\left( x \right) \right)=kd\left( g\left( x \right) \right).\] So, in our case f(x) = – 3x.
\[\dfrac{d\left( -3x \right)}{dx}=-3\dfrac{dx}{dx}=-3\]
As \[\dfrac{dx}{dx}=1.\]
So, the derivative of f(x) = – 3x is – 3.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

The chemical formula of tear gas is A CO Cl 2 B C 10 class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

