Question

Find the gradient of the curve y = x^5/2 at the point where x = 4 ​

Answer 1

Given:

The equation of the curve is:

$y=x^{\frac{5}{2}}$

To find:

The gradient of the given curve at the point where $x = 4$.

Solution:

We have,

$y=x^{\frac{5}{2}}$

Differentiate with respect to x.

$y'=\dfrac{5}{2}x^{\frac{5}{2}-1}$       $[\because \dfrac{d}{dx}x^n=nx^{n-1}]$

$y'=\dfrac{5}{2}x^{\frac{3}{2}}$

Substituting $x=4$, we get

$y'=\dfrac{5}{2}(4)^{\frac{3}{2}}$

$y'=\dfrac{5}{2}(2^2)^{\frac{3}{2}}$

Using properties of exponents, we get

$y'=\dfrac{5}{2}(2)^{2\times \frac{3}{2}}$         $[\because (a^m)^n=a^{mn}]$

$y'=\dfrac{5}{2}(2)^{3}$

$y'=\dfrac{5}{2}(8)$

$y'=20$

Therefore, the gradient of the given curve at the point where $x = 4$ is 20.