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Question: How do you use the definition of a derivative to find the derivative of \(G\left( t \right)=\dfrac{1...

How do you use the definition of a derivative to find the derivative of G(t)=1−6t5+t?G\left( t \right)=\dfrac{1-6t}{5+t}?

Explanation

Solution

The derivative is defined in terms of limits. Let FF be a function defined in an interval. Then the derivative of FF at a point xx in the interval is defined as F′(x)=lim⁡h→0 F(x+h)−F(x)h.{F}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{F\left( x+h \right)-F\left( x \right)}{h}.

Complete step by step solution:
Let us consider the given function G(t)=1−6t5+t.G\left( t \right)=\dfrac{1-6t}{5+t}.
We are asked to find the derivative of the given function using the definition of derivative. We know that the derivative is defined in terms of limit.
Suppose that FF is a function defined in an interval [a,b].\left[ a,b \right]. Then the derivative of the function FF at a point xx in the interval [a,b]\left[ a,b \right] is defined as F′(x)=lim⁡h→0 F(x+h)−F(x)h.{F}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{F\left( x+h \right)-F\left( x \right)}{h}.
Let us apply this equation in our problem. We will get G′(t)=lim⁡h→0 G(t+h)−G(t)h.{G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{G\left( t+h \right)-G\left( t \right)}{h}.
So, we will get G(t+h)=1−6(t+h)5+(t+h).G\left( t+h \right)=\dfrac{1-6\left( t+h \right)}{5+\left( t+h \right)}.
This will give us G′(t)=1−6t−6h5+t+h.{G}'\left( t \right)=\dfrac{1-6t-6h}{5+t+h}.
Therefore, the derivative of the given function can be obtained by
⇒G′(t)=lim⁡h→0 (1−6t−6h)5+t+h−1−6t5+th.\Rightarrow {G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\left( 1-6t-6h \right)}{5+t+h}-\dfrac{1-6t}{5+t}}{h}.
And, from this, when we will take LCM, we will get
⇒G′(t)=lim⁡h→0 (1−6t−6h)(5+t)−(1−6t)(5+t+h)(5+t+h)(5+t)h.\Rightarrow {G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\left( 1-6t-6h \right)\left( 5+t \right)-\left( 1-6t \right)\left( 5+t+h \right)}{\left( 5+t+h \right)\left( 5+t \right)}}{h}.
We will get G′(t)=lim⁡h→0 (5−30t−30h+t−6t2−6th)−(5+t+h−30t−6t2−6th)h(5+t+h)(5+t).{G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 5-30t-30h+t-6{{t}^{2}}-6th \right)-\left( 5+t+h-30t-6{{t}^{2}}-6th \right)}{h\left( 5+t+h \right)\left( 5+t \right)}.
And this will give us G′(t)=lim⁡h→0 5−30t−30h+t−6t2−6ht−5−t−h+30t+6t2+6thh(5+t+h)(5+t).{G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{5-30t-30h+t-6{{t}^{2}}-6ht-5-t-h+30t+6{{t}^{2}}+6th}{h\left( 5+t+h \right)\left( 5+t \right)}.
From this, after cancelling off the terms with opposite signs, we will obtain
⇒G′(t)=lim⁡h→0 −31hh(5+t+h)(5+t).\Rightarrow {G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-31h}{h\left( 5+t+h \right)\left( 5+t \right)}.
Now, we will cancel hh off the numerator and the denominator to get
⇒G′(t)=lim⁡h→0 −31(5+t+h)(5+t).\Rightarrow {G}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-31}{\left( 5+t+h \right)\left( 5+t \right)}.
When h→0,h\to 0, the derivative will become G′(t)=−31(5+t+0)(5+t).{G}'\left( t \right)=\dfrac{-31}{\left( 5+t+0 \right)\left( 5+t \right)}.
That is G′(t)=−31(5+t)(5+t)=−31(5+t)2.{G}'\left( t \right)=\dfrac{-31}{\left( 5+t \right)\left( 5+t \right)}=\dfrac{-31}{{{\left( 5+t \right)}^{2}}}.
Hence the derivative is G′(t)=−31(5+t)2.{G}'\left( t \right)=\dfrac{-31}{{{\left( 5+t \right)}^{2}}}.

Note: When we differentiate a function that is a fraction, we will use the quotient rule of differentiation. It states that the derivative if a function uv\dfrac{u}{v} can be calculated by ddxuv=vdudx−udvdxv2\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} where uu and vv are the functions of x.x. We will get G′(t)=(5+t)ddt(1−6t)−(1−6t)ddt(5+t)(5+t)2.{G}'\left( t \right)=\dfrac{\left( 5+t \right)\dfrac{d}{dt}\left( 1-6t \right)-\left( 1-6t \right)\dfrac{d}{dt}\left( 5+t \right)}{{{\left( 5+t \right)}^{2}}}. After solving this, we will get the derivative.