Ex 5.5 Q1 To Q10 || lec 8 || Continuity and Differentiability || NCERT || Class 12 | Chapter 5 ||

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[vanshsharmav]

hw answer is done but i m unable to write it here

[technogenix.]

To differentiate the expression \(5(n+3)^2 \cdot (n+4)^3 \cdot (n+5)^4\) with respect to \(x\), you can use the product rule of differentiation. The product rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by:

\[(u \cdot v)' = u' \cdot v + u \cdot v'\]

In your case, you have three functions: \(u(x) = 5(n+3)^2\), \(v(x) = (n+4)^3\), and \(w(x) = (n+5)^4\).

Now, let's differentiate each of these functions with respect to \(n\) first:

1. \(u'(n) = 2 \cdot 5(n+3)\cdot \frac{d}{dn}(n+3) = 10(n+3)\)
2. \(v'(n) = 3(n+4)^2 \cdot \frac{d}{dn}(n+4) = 3(n+4)^2\)
3. \(w'(n) = 4(n+5)^3 \cdot \frac{d}{dn}(n+5) = 4(n+5)^3\)

Now, apply the product rule:

\[
\begin{align*}
\frac{d}{dx}\left(5(n+3)^2 \cdot (n+4)^3 \cdot (n+5)^4\right) &= 5 \cdot 10(n+3) \cdot (n+4)^3 \cdot (n+5)^4 \\
&+ 5(n+3)^2 \cdot 3(n+4)^2 \cdot (n+5)^4 \\
&+ 5(n+3)^2 \cdot (n+4)^3 \cdot 4(n+5)^3
\end{align*}
\]

Simplify each term and you'll have the derivative of the expression with respect to \(x\):

\[
\frac{d}{dx}\left(5(n+3)^2 \cdot (n+4)^3 \cdot (n+5)^4\right) = 50(n+3)(n+4)^3(n+5)^4 + 15(n+3)^2(n+4)^2(n+5)^4 + 20(n+3)^2(n+4)^3(n+5)^3
\]

This is the derivative of the given expression with respect to \(x\).