Question

What is the slope of `f(x)=-xe^(x-3x^3)` at `x=-1`?

Answer 1

The reqd. slope `=[dy/dx]_(x=-1)=f'(-1)=-9e^2.`

Explanation:

Slope of a curve `: y=f(x)` at a particular pt. is, by defn., the value of `dy/dx` at that pt.

Here, `y=f(x)= -x*e^(x-3x^3).`

`:. dy/dx=f'(x)=-{x*d/dx(e^(x-3x^3))+e^(x-3x^3)*d/dx(x)}=-{x*e^(x-3x^3)*d/dx((x-3x^3))+e^(x-3x^3)*1}=-{x*e^(x-3x^3)*(1-9x^2)+e^(x-3x^3)}=-e^(x-3x^3){x-9x^3+1}.`

Therefore, the reqd. slope `=[dy/dx]_(x=-1)=f'(-1)=-e^(-1+3)(-1+9+1)=-9e^2`