Question

Let f, g and h are differentiable functions. If f (0) = 1, g (0) = 2, h (0) = 3 and the derivatives of their pair wise products at x = 0 are \[{{\left( fg \right)}^{'}}\left( 0 \right)=6;{{\left( gh \right)}^{'}}\left( 0 \right)=4;{{\left( hf \right)}^{'}}\left( 0 \right)=5\] then compute the value of (fgh)’(0).

Answer 1

Hint: First, find the expression of the derivative of the product of three functions u, v, w given by the product rule as \[\dfrac{d\left( uvw \right)}{dx}=uv\dfrac{dw}{dx}+uw\dfrac{dv}{dx}+vw\dfrac{du}{dx}.\] Similarly, using the above formula, find the expression of the derivative of the product of two functions. Substitute the value of x = 0 in all the places and use the given relations to compute the value of (fgh)’(0).

Complete step by step answer:
Here, we have been given three functions f, g and h and the values of these functions at x = 0 is given. Also, we have been provided with the derivative of their product, taken two at a time, at x = 0. We have been asked to determine the value of the derivative of the product of f, g and h at x = 0. Now, we know that if we have three functions u, v and w then their derivative of the product is given by the product rule of differentiation. So, we have,
\[\dfrac{d\left( uvw \right)}{dx}=uv\dfrac{dw}{dx}+uw\dfrac{dv}{dx}+vw\dfrac{du}{dx}\]
Therefore, the derivative of the product of f, h and h will be given as
\[\Rightarrow \dfrac{d\left( fgh \right)}{dx}=fg\dfrac{dh}{dx}+fh\dfrac{dg}{dx}+gh\dfrac{df}{dx}\]
\[\Rightarrow \dfrac{d\left( fgh \right)}{dx}=fg{{h}^{'}}\left( x \right)+fh{{g}^{'}}\left( x \right)+gh{{f}^{'}}\left( x \right)\]
\[\Rightarrow \dfrac{d\left( fgh \right)}{dx}=f\left( 0 \right).g\left( 0 \right).{{h}^{'}}\left( 0 \right)+f\left( 0 \right).h\left( 0 \right).{{g}^{'}}\left( 0 \right)+g\left( 0 \right).h\left( 0 \right).{{f}^{'}}\left( 0 \right)\]
Substituting the provided values of f(0), g(0) and h(0) in the above relation, we get,
\[\Rightarrow \dfrac{d\left( fgh \right)}{dx}=2{{h}^{'}}\left( 0 \right)+3{{g}^{'}}\left( 0 \right)+6{{f}^{'}}\left( 0 \right)......\left( i \right)\]
Now we have, using the product rule,
\[\dfrac{d\left( fg \right)}{dx}=g\dfrac{df}{dx}+f\dfrac{dg}{dx}\]
\[\Rightarrow \dfrac{d\left( fg \right)}{dx}=g{{f}^{'}}\left( x \right)+f{{g}^{'}}\left( x \right)\]
\[\Rightarrow {{\left( fg \right)}^{'}}\left( x \right)=g{{f}^{'}}\left( x \right)+f{{g}^{'}}\left( x \right)\]
Substituting x = 0, we get,
\[\Rightarrow {{\left( fg \right)}^{'}}\left( 0 \right)=g\left( 0 \right){{f}^{'}}\left( 0 \right)+f\left( 0 \right){{g}^{'}}\left( 0 \right)\]
\[\Rightarrow 6=2{{f}^{'}}\left( 0 \right)+{{g}^{'}}\left( 0 \right)......\left( ii \right)\]
Similarly, using the product rule, we can write,
\[\Rightarrow {{\left( gh \right)}^{'}}\left( 0 \right)=g\left( 0 \right){{h}^{'}}\left( 0 \right)+h\left( 0 \right){{g}^{'}}\left( 0 \right)\]
\[\Rightarrow 4=2{{h}^{'}}\left( 0 \right)+3{{g}^{'}}\left( 0 \right).....\left( iii \right)\]
Also, we have,
\[\Rightarrow {{\left( fh \right)}^{'}}\left( 0 \right)=f\left( 0 \right){{h}^{'}}\left( 0 \right)+h\left( 0 \right){{f}^{'}}\left( 0 \right)\]
\[\Rightarrow 5={{h}^{'}}\left( 0 \right)+3{{f}^{'}}\left( 0 \right)......\left( iv \right)\]
Solving the equations (ii), (iii) and (iv), we get,
\[\Rightarrow {{f}^{'}}\left( 0 \right)=2;{{g}^{'}}\left( 0 \right)=2;{{h}^{'}}\left( 0 \right)=-1\]
Substituting these values in equation (i), we get,
\[\Rightarrow \dfrac{d\left( fgh \right)}{dx}=2\times \left( -1 \right)+3\times 2+6\times 2\]
\[\Rightarrow {{\left( fgh \right)}^{'}}\left( 0 \right)=-2+6+12\]
\[\Rightarrow {{\left( fgh \right)}^{'}}\left( 0 \right)=16\]

Note:
One may note that without memorizing the product rule you cannot solve this question. Do not get confused between the formulas of differentiation of products of two functions and three functions. We have to differentiate one function and keep the other functions as products at one time. This is the basic rule and is applied even for the product rule of the product of ‘n’ given functions.