Answer to Question #212388 in Calculus for Kenneth

Question

1. Find the derivative of the following functions with respect to x.

(a) f(x) = x^-1/2 minus xe^x + 10^x + e^x^2 / 2x minus cube root of 15

(b) F(x) = e^x sinx + sec x + ln x^2

2. If y^5 + 4xy^3 + x^3 +1 =0, find dy / dx and determine its values at the point ( x, y) = ( 0, -1).

Answer 1

1.

(a) $f(x)=x^\frac{-1}{ 2}-xe^x+10^x+\frac{e^{x^2}}{2x}-\sqrt[3]{15};$

$\frac{df}{dx}=-\frac{1}{2}x^\frac{-3}{2}-(x+1)e^x+10^xln10+\frac{4x^2e^{x^2}-2e^{x^2}}{4x^2}=$

$=-\frac{1}{2}x^\frac{-3}{2}-(x+1)e^x+10^xln10+\frac{(2x^2-1)e^{x^2}}{2x^2}$;

(b) $F(x)=e^xsinx+secx+lnx^2;$

$\frac{dF}{dx}=e^xsinx+e^xcosx+\frac{tanx}{cosx}+\frac{2}{x}=$

$=(sinx+cosx)e^x+\frac{tanx}{cosx}+\frac{2}{x};$

2.

We use the implicit function differentiation rule:

$5y^4\frac{dy}{dx}+4y^3+12xy^2\frac{dy}{dx}+3x^2=0;$

$\frac{dy}{dx}(5y^4+12xy^2)=-(4y^3+3x^2);$

$\frac{dy}{dx}=-\frac{4y^3+3x^2}{5y^4+12xy^2} .$

At the point (x,y)=(0,-1) we have:

$\frac{y}{dx}=-\frac{4\cdot (-1)^3+0}{5\cdot (-1)^4+0}=\frac{4}{5} .$