How do you differentiate \[y={{e}^{2{{x}^{3}}}}\] ?
Answer
616.5k+ views
Hint: Now we know that the given function is a composite function of the form $f\left( g\left( x \right) \right)$ . Hence to differentiate the function we will use the chain rule. Now we know that the chain rule of for a composite function is given as $f'\left( g\left( x \right) \right)g'\left( x \right)$ here we have $f\left( x \right)={{e}^{x}}$ and $g\left( x \right)=2{{x}^{3}}$ . Now we know that differentiation of ${{e}^{x}}$ is ${{e}^{x}}$ and differentiation of ${{x}^{n}}=n{{x}^{n-1}}$ Hence we will use the chain rule and find the differentiation of the function.
Complete step-by-step answer:
Now we are given a function \[y={{e}^{2{{x}^{3}}}}\] .
We know that the given function is a composite function.
Composite functions are nothing but a function inside a function, Hence they are functions of the form $f\left( g\left( x \right) \right)$ .
To differentiate a composite function we use chain rule.
Let us first understand the concept of chain rule.
Now let us say we have a composite function $f\left( g\left( x \right) \right)$ .
Then the differentiation of the function is given by $f'\left( g\left( x \right) \right)g'\left( x \right)$ .
Now consider the given function \[y={{e}^{2{{x}^{3}}}}\].
Here $f\left( x \right)={{e}^{x}}$ and $g\left( x \right)=2{{x}^{3}}$
Now we know that differentiation of ${{e}^{x}}$ is ${{e}^{x}}$ .
Hence we have $f'\left( x \right)={{e}^{x}}$ .
Hence $f'\left( g\left( x \right) \right)={{e}^{2{{x}^{3}}}}$ .
Now similarly $g\left( x \right)=2{{x}^{3}}$
Now we know that $f'\left( cx \right)=cf'\left( x \right)$ and differentiation of ${{x}^{n}}=n{{x}^{n-1}}$
Hence the differentiation of $2{{x}^{3}}=2\left( 3 \right){{x}^{3-2}}=6{{x}^{2}}$
Hence we have $g'\left( x \right)=6{{x}^{2}}$
Hence the differentiation of \[y={{e}^{2{{x}^{3}}}}\] by chain rule is given by ${{e}^{2{{x}^{3}}}}\left( 6{{x}^{2}} \right)$
Hence we have the differentiation of the given function.
Note: Now note that composite functions are functions inside a function. Not to be confused with multiplication of functions. Composite functions can be written as $f\left( g\left( x \right) \right)$ and the differentiation is given by $f'\left( g\left( x \right) \right)g'\left( x \right)$ and the multiplication of functions is written as $f\left( x \right).g\left( x \right)$ and the differentiation is given as $f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)$ .
Complete step-by-step answer:
Now we are given a function \[y={{e}^{2{{x}^{3}}}}\] .
We know that the given function is a composite function.
Composite functions are nothing but a function inside a function, Hence they are functions of the form $f\left( g\left( x \right) \right)$ .
To differentiate a composite function we use chain rule.
Let us first understand the concept of chain rule.
Now let us say we have a composite function $f\left( g\left( x \right) \right)$ .
Then the differentiation of the function is given by $f'\left( g\left( x \right) \right)g'\left( x \right)$ .
Now consider the given function \[y={{e}^{2{{x}^{3}}}}\].
Here $f\left( x \right)={{e}^{x}}$ and $g\left( x \right)=2{{x}^{3}}$
Now we know that differentiation of ${{e}^{x}}$ is ${{e}^{x}}$ .
Hence we have $f'\left( x \right)={{e}^{x}}$ .
Hence $f'\left( g\left( x \right) \right)={{e}^{2{{x}^{3}}}}$ .
Now similarly $g\left( x \right)=2{{x}^{3}}$
Now we know that $f'\left( cx \right)=cf'\left( x \right)$ and differentiation of ${{x}^{n}}=n{{x}^{n-1}}$
Hence the differentiation of $2{{x}^{3}}=2\left( 3 \right){{x}^{3-2}}=6{{x}^{2}}$
Hence we have $g'\left( x \right)=6{{x}^{2}}$
Hence the differentiation of \[y={{e}^{2{{x}^{3}}}}\] by chain rule is given by ${{e}^{2{{x}^{3}}}}\left( 6{{x}^{2}} \right)$
Hence we have the differentiation of the given function.
Note: Now note that composite functions are functions inside a function. Not to be confused with multiplication of functions. Composite functions can be written as $f\left( g\left( x \right) \right)$ and the differentiation is given by $f'\left( g\left( x \right) \right)g'\left( x \right)$ and the multiplication of functions is written as $f\left( x \right).g\left( x \right)$ and the differentiation is given as $f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)$ .
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