What is the slope of the tangent line to y equals x3 at x equals 4?

Answer 1

Do you mean y=3x or y=x3?

I'll assume it's the latter.

The first method to solving this is the easiest, the chain rule.

Multiply the coefficient by the value of the exponent and reduce the exponent by 1. (f(x)=nAxn-1)

You get y=3x2

Therefore, f(4)=3(42)

f(4)=48

The longer method of solving this goes as follows:

Your tangent formula is f'(x)=limh->0(f(x+h)-f(x))/h

We know that f(x)=x3 so f(x+h)=(x+h)3

When we put this in the formula we have:

f'(x)=limh->0((x+h)3-x3)/h

f'(x)=limh->0((x+h)(x+h)(x+h)-x3)/h

f'(x)=limh->0((x2+2hx+h2)(x+h)-x3)/h

f'(x)=limh->0(x3+x2h+2x2h+2xh2+xh2+h3-x3)/h

f'(x)=limh->0(3x2+3xh+h2)

f'(x)=3x2+3x(0)+(02)

f'(x)=3x2

And from there again we sub in 4 for x.

f'(4)=3(42)

f'(4)=48