Find the value of $r'(1)$, if $r(x) = f(g(h(x)))$, where $h(1) = 2$, $g(2) = 3$, $h'(1) = 4$, $g'(2) = 5$ and $f'(3) = 6$.
Answer
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Hint:In this problem we have given that $r(x)$ is a composite function as it involves two other functions as well and to find the value of $r'(1)$, we need to solve this by chain rule as the values of the functions are known at certain points. Chain rule is a formula to find the derivative of a composite function or we can say that it is used to differentiate composite functions.
Formula Used:
Chain rule for two functions is-
$ \Rightarrow \dfrac{d}{{dx}}(f(g(x))) = f'(g(x)) \times g'(x)$
Where, f is the outside function and g is inside the function.
Chain rule applied in this question is-
$ \Rightarrow \dfrac{d}{{dx}}(f(g(h(x)))) = h'(x) \times g'(h(x)) \times f'(g(h(x))$
Complete step by step answer:
Since, we are to deal with composite functions. It is given that $r(x) = f(g(h(x)))$ and we also know the values at certain points i.e. $h(1) = 2$, $g(2) = 3$,$h'(1) = 4$,$g'(2) = 5$ and $f'(3) = 6$. The chain rule for two functions is,
$\dfrac{d}{{dx}}(f(g(x))) = f'(g(x)) \times g'(x)$
And the chain rule for $f(g(h(x)))$ is,
$\dfrac{d}{{dx}}(f(g(h(x)))) = h'(x) \times g'(h(x)) \times f'(g(h(x))$
Now, we will substitute the values given in the question into this chain rule.
$r'(x) = h'(x) \times g'(h(x)) \times f'(g(h(x))) \\
\Rightarrow r'(1) = h'(1) \times g'(h(1)) \times f'(g(h(1))) \\
\Rightarrow r'(1) = 4 \times g'(2) \times f'(g(2) \\
\Rightarrow r'(1) = 4 \times 5 \times f'(3) \\
\Rightarrow r'(1) = 4 \times 5 \times 6 $
On further solving, we get,
$\therefore r'(1) = 120$
Hence, the value of $r'(1)$ is $120$.
Note:The chain rule should be known to us for solving these type of questions and in this question there is not any other function given, what we have given is the values of the functions at certain points and when the chain rule is applied we need to substitute the values and after substituting, if we get another function then the another value of that function must be substituted until we get our answer.
Formula Used:
Chain rule for two functions is-
$ \Rightarrow \dfrac{d}{{dx}}(f(g(x))) = f'(g(x)) \times g'(x)$
Where, f is the outside function and g is inside the function.
Chain rule applied in this question is-
$ \Rightarrow \dfrac{d}{{dx}}(f(g(h(x)))) = h'(x) \times g'(h(x)) \times f'(g(h(x))$
Complete step by step answer:
Since, we are to deal with composite functions. It is given that $r(x) = f(g(h(x)))$ and we also know the values at certain points i.e. $h(1) = 2$, $g(2) = 3$,$h'(1) = 4$,$g'(2) = 5$ and $f'(3) = 6$. The chain rule for two functions is,
$\dfrac{d}{{dx}}(f(g(x))) = f'(g(x)) \times g'(x)$
And the chain rule for $f(g(h(x)))$ is,
$\dfrac{d}{{dx}}(f(g(h(x)))) = h'(x) \times g'(h(x)) \times f'(g(h(x))$
Now, we will substitute the values given in the question into this chain rule.
$r'(x) = h'(x) \times g'(h(x)) \times f'(g(h(x))) \\
\Rightarrow r'(1) = h'(1) \times g'(h(1)) \times f'(g(h(1))) \\
\Rightarrow r'(1) = 4 \times g'(2) \times f'(g(2) \\
\Rightarrow r'(1) = 4 \times 5 \times f'(3) \\
\Rightarrow r'(1) = 4 \times 5 \times 6 $
On further solving, we get,
$\therefore r'(1) = 120$
Hence, the value of $r'(1)$ is $120$.
Note:The chain rule should be known to us for solving these type of questions and in this question there is not any other function given, what we have given is the values of the functions at certain points and when the chain rule is applied we need to substitute the values and after substituting, if we get another function then the another value of that function must be substituted until we get our answer.
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