$\displaystyle y=\frac{4}{x^2}$ why is this $\displaystyle y=\frac{4x}{x^3}$ and not $\displaystyle y=\frac{4x}{0.3333x^3}$ sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third
Follow Math Help Forum on Facebook and Google+
Originally Posted by coyoteflare $\displaystyle y=\frac{4}{x^2}$ why is this $\displaystyle y=\frac{4x}{x^3}$ and not $\displaystyle y=\frac{4x}{0.3333x^3}$ sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third Are you trying to integrate $\displaystyle \frac{4}{x^2}$ i.e. $\displaystyle \int \frac{4}{x^2}\, dx$?
Originally Posted by danny arrigo Are you trying to integrate $\displaystyle \frac{4}{x^2}$ i.e. $\displaystyle \int \frac{4}{x^2}\, dx$? heres the question A is the point (2,1) on the curve $\displaystyle y=\frac{4}{x^2}$ use calculas to find the gradient of the curve at A
Originally Posted by coyoteflare heres the question A is the point (2,1) on the curve $\displaystyle y=\frac{4}{x^2}$ use calculas to find the gradient of the curve at A Oh, you want the slope of the tangent. In this case you'll need the derivative. So $\displaystyle \frac{d}{dx} \left( \frac{4}{x^2} \right) = \frac{d}{dx} \left( 4 x^{-2} \right)$ Once you have this, evaluate it at $\displaystyle x = 2$ and you'll have your answer.
View Tag Cloud