Consider :- \( \scriptsize y = \normalsize \frac{1}{(3x\; – \;4)^2 }\scriptsize = (3x -4)^{-2}\)
A function such as this sets a problem for differentiation.
It cannot be expressed as a polynomial. So we look at a very important technique devised to handle such functions.
To find \( \frac{dy}{dx} \) for the function above, we treat \( \scriptsize = (3x -4)^{-2} \! \! \! \) as a composite function built up in two stages from a core function (3x – 4), which we call u and then taking \( \scriptsize y = u^{-2} \) So we have u = (3x – 4)
Hence the general form of a function of function rule also called chain rule which is given as
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
Then in the above problem,
\( \scriptsize y = (3x -4)^{-2} \)
Solution
\( \scriptsize y = u^{-2}, u = (3x -4) \)Thus \( \frac{dy}{du} = \scriptsize -2u^{-3} \normalsize = \frac{-2}{u^3} \)
while \( \frac{du}{dx} \scriptsize = 3\)
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\) \( \frac{dy}{dx} = \; – \frac{2}{u^3} \times \scriptsize 3 =\; – \normalsize\frac{6}{u^3} \)Substituting for u
= \(– \frac{6}{(3x \;-\; 4)^3} \)
Example:
2. Differentiate \( \scriptsize y = (4x -5)^6 \)
Let \( \scriptsize y = u^{6}, u = (4x – 5) \)
Thus \( \frac{dy}{du} = \scriptsize 6u^{5}\)
while \( \frac{du}{dx} \scriptsize = 4\)
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)= \(\scriptsize 6u^{5} \times 4 \)
\( \frac{dy}{dx} = \scriptsize 24u^{5} \)and substituting u = 4x – 5 gives \( \frac{dy}{dx} = \scriptsize 24(4x-5)^{5} \)
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