Find the tangent to the curve y=5x⁴-3x²+2x-1 at x=1

The correct answer is 16.

To find the slope of the tangent to the curve at a given point, we differentiate the equation of the curve with respect to $x$ to find the derivative, and then evaluate this derivative at the given value of $x$.

The equation of the curve is:

$$y=5x^4-3x^2+2x-1$$

We differentiate $y$ with respect to $x$ using the power rule of differentiation, which states that for any real number $n$:

$$\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$$

Applying the power rule to each term of the curve's equation, we get:

$$\frac{dy}{dx}=\frac{d}{dx}\left(5x^4-3x^2+2x-1\right)$$

$$\frac{dy}{dx}=5\cdot \left(4x^3\right)-3\cdot \left(2x\right)+2\cdot \left(1\right)-0$$

Simplifying the expression for the derivative:

$$\frac{dy}{dx}=20x^3-6x+2$$

To find the slope of the tangent at $x=1$, we substitute $x=1$ into the derivative function:

$${\left.\frac{dy}{dx}\right|}_{x=1}=20{\left(1\right)}^3-6\left(1\right)+2$$

$${\left.\frac{dy}{dx}\right|}_{x=1}=20-6+2$$

$${\left.\frac{dy}{dx}\right|}_{x=1}=16$$

Therefore, the slope of the tangent to the curve at the point where $x=1$ is 16.