Question

If \(g\left( x \right) = {x^4} - 2\), find \(g'\left( 1 \right)\) and use it to find an equation of the tangent line to the curve \(y = {x^4} - 2\) at the point \(\left( {1, - 1} \right)\).

Short answer

The equation of the tangent line to the curve \(y = {x^4} - 2\) at a point \(\left( {1, - 1} \right)\) is \(y = 4x - 5\).

Step-by-step solution

The Derivative of a function

The formula of the derivative of a function is shown below:

\(f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\)

The derivative of the function at \(x = 1\)

Substitute the value \(a = 1\) in the above formula and solve as follows:

\(\begin{aligned}f'\left( 1 \right) &= \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( x \right)}^4} - 2 - \left( {{1^4} - 2} \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 2 + 1}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( 1 \right) &= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}{{x - 1}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)}}\\ &= \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + 1} \right)\left( {x + 1} \right)\\ &= \left( {1 + 1} \right)\left( {1 + 1} \right)\\ &= 4\end{aligned}\)

The slope of the tangent line to the curve

Recall the fact the derivative \(f'\left( a \right)\) of a function \(y = f\left( x \right)\) is equal to the slope of the tangent line to the curve \(y\) at the point \(x = a\).

Therefore, the slope of the tangent line to the given curve at \(x = 1\) is \(g'\left( 1 \right) = 4\).

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

Equation of the tangent line at point \(\left( {1, - 1} \right)\)

Substitute \(\left( {{x_1},{y_1}} \right) = \left( {1, - 1} \right)\) and \(m = 4\) in the equation of the line and solve as follows:

\(\begin{aligned}y - {y_1} &= m\left( {x - {x_2}} \right)\\y - \left( { - 1} \right) &= 4\left( {x - 1} \right)\\y + 1 &= 4x - 4\\y &= 4x - 5\end{aligned}\)

Thus, the equation of the tangent line to the curve \(y = {x^4} - 2\) at a point \(\left( {1, - 1} \right)\) is \(y = 4x - 5\).