Question

If $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ where $f\left( 1 \right) = 7$ and $f'\left( 1 \right) = 4$ , how do you find $h'\left( 1 \right)$ ?

Answer 1

Hint : In the given problem, we are required to differentiate $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ with respect to x. Since, $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ is a composite function, we will have to apply chain rule of differentiation in the process of differentiation. So, differentiation of $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ with respect to x will be done layer by layer using the chain rule of differentiation.

Complete step-by-step answer :
To find derivative of $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ with respect to $x$ we have to find differentiate $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$ with respect to $x$ .
So, $h\left( x \right) = \sqrt {4 + 3f\left( x \right)}$
Differentiating both sides with respect to x, we get,
$h\left( x \right) = \dfrac{d}{{dx}}\left( {\sqrt {4 + 3f\left( x \right)} } \right)$
Now, we have to apply chain rule in order to differentiate the right side of the equation.
We know that according to chain rule of differentiation,
$\dfrac{d}{{dx}}h\left[ {g\left( {f\left( x \right)} \right)} \right] = h'\left[ {g\left( {f\left( x \right)} \right)} \right]g'\left( {f\left( x \right)} \right)f'\left( x \right)$
So, we get,
$\Rightarrow h'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{3f'\left( x \right)}}{{\sqrt {4 + 3f\left( x \right)} }}} \right)$
Putting the value of x as $1$ , we get,
$\Rightarrow h'\left( 1 \right) = \dfrac{1}{2}\left( {\dfrac{{3f'\left( 1 \right)}}{{\sqrt {4 + 3f\left( 1 \right)} }}} \right)$
Substituting the values of $f\left( 1 \right)$ and $f'\left( 1 \right)$ as given in the question, we get,
$\Rightarrow h'\left( 1 \right) = \dfrac{1}{2}\left( {\dfrac{{3 \times 4}}{{\sqrt {4 + 3 \times 7} }}} \right)$
Simplifying further and doing the calculations,
$h'\left( 1 \right) = \dfrac{1}{2}\left( {\dfrac{{12}}{{\sqrt {25} }}} \right)$
We know that the square root of $25$ is $5$ . So, putting the value in the expression, we get,
$h'\left( 1 \right) = \dfrac{1}{2}\left( {\dfrac{{12}}{5}} \right)$
Cancelling the common terms in numerator and denominator, we get,
$h'\left( 1 \right) = \left( {\dfrac{6}{5}} \right)$
So, the correct answer is “$h'\left( 1 \right) = \left( {\dfrac{6}{5}} \right)$”.

Note : The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.