Question

examples: differentiate the following in lagr

1. $f(x)=(x^{3}-3)^{5}(2x^{2}-7)^{3}$

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u=(x^{3}-3)^{5}\) and \(v=(2x^{2}-7)^{3}\).
First, find \(u^\prime\) using the chain rule. If \(u = g(h(x))\) where \(g(x)=x^{5}\) and \(h(x)=x^{3}-3\), then \(u^\prime = g^\prime(h(x))\cdot h^\prime(x)\). \(g^\prime(x) = 5x^{4}\), \(h^\prime(x)=3x^{2}\), so \(u^\prime=5(x^{3}-3)^{4}\cdot3x^{2}=15x^{2}(x^{3}-3)^{4}\).
Next, find \(v^\prime\) using the chain rule. If \(v = m(n(x))\) where \(m(x)=x^{3}\) and \(n(x)=2x^{2}-7\), then \(v^\prime = m^\prime(n(x))\cdot n^\prime(x)\). \(m^\prime(x) = 3x^{2}\), \(n^\prime(x)=4x\), so \(v^\prime=3(2x^{2}-7)^{2}\cdot4x = 12x(2x^{2}-7)^{2}\).

Step2: Substitute into the product rule

\(f^\prime(x)=u^\prime v+uv^\prime\)
\(f^\prime(x)=15x^{2}(x^{3}-3)^{4}(2x^{2}-7)^{3}+(x^{3}-3)^{5}\cdot12x(2x^{2}-7)^{2}\)
Factor out common terms:
\(f^\prime(x)=3x(x^{3}-3)^{4}(2x^{2}-7)^{2}[5x(2x^{2}-7)+4(x^{3}-3)]\)
Expand the brackets inside the square - brackets:
\(5x(2x^{2}-7)+4(x^{3}-3)=10x^{3}-35x + 4x^{3}-12=14x^{3}-35x - 12\)

Answer:

\(f^\prime(x)=3x(x^{3}-3)^{4}(2x^{2}-7)^{2}(14x^{3}-35x - 12)\)