Question

find an equation for the tangent line to the graph of ( y = left(x^3 - 25x
ight)^{10} ) at the point ( (-5, 0) ). the equation of the tangent line is ( y = square ). (simplify your answer.)

Explanation:

Step1: Find the derivative of \( y=(x^3 - 25x)^{10} \) using the chain rule.

The chain rule states that if \( y = u^{n} \) and \( u = f(x) \), then \( y'=n\cdot u^{n - 1}\cdot u' \). Let \( u=x^3 - 25x \) and \( n = 10 \). First, find \( u' \): \( u'=\frac{d}{dx}(x^3-25x)=3x^2 - 25 \). Then, by the chain rule, \( y'=10(x^3 - 25x)^{9}\cdot(3x^2 - 25) \).

Step2: Evaluate the derivative at \( x=-5 \) to find the slope of the tangent line.

Substitute \( x = - 5 \) into \( y' \):
First, calculate \( u(-5)=(-5)^3-25\times(-5)=-125 + 125 = 0 \).
Then, \( 3x^2-25\) at \( x=-5 \) is \( 3\times(-5)^2-25=3\times25 - 25=75 - 25 = 50 \).
So \( y'(-5)=10\times(0)^{9}\times50 = 0 \). The slope \( m \) of the tangent line is \( 0 \).

Step3: Use the point - slope form of a line to find the equation of the tangent line.

The point - slope form is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(-5,0) \) and \( m = 0 \).
Substitute these values into the formula: \( y-0=0\times(x + 5) \), which simplifies to \( y = 0 \).

Answer:

\( y = 0 \)