Question

Find the slope of the tangent to the curve $f\left(x\right)=2x^6+x^4-1$ at x=1.

Answer 1

Slope of the tangent at a point x=a is the value of the derivative ay x=a

We have,

$f\left(x\right)=2x^6+x^4-1$
$=\frac{d}{dx}(2x^6+x^4-1)$

$=2\frac{d{x}^6}{dx}+\frac{d{x}^4}{dx}-\frac{d.1}{dx}$

$=12x^5+4x^3-0$

$=12x^5+4x^3$

$\frac{dy}{dx}$ at $x=1$

$12(1{)}^5+4(1{)}^3$

=12+4

=16

$\therefore$ The slope of the tangent to the curve
$f\left(x\right)=2x^6+x^4-1$ at $x=1$ is 16.