1.
(a) f ( x ) = x − 1 2 − x e x + 1 0 x + e x 2 2 x − 15 3 ; f(x)=x^\frac{-1}{ 2}-xe^x+10^x+\frac{e^{x^2}}{2x}-\sqrt[3]{15}; f ( x ) = x 2 − 1 − x e x + 1 0 x + 2 x e x 2 − 3 15 ;
d f d x = − 1 2 x − 3 2 − ( x + 1 ) e x + 1 0 x l n 10 + 4 x 2 e x 2 − 2 e x 2 4 x 2 = \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}= d x df = − 2 1 x 2 − 3 − ( x + 1 ) e x + 1 0 x l n 10 + 4 x 2 4 x 2 e x 2 − 2 e x 2 =
= − 1 2 x − 3 2 − ( x + 1 ) e x + 1 0 x l n 10 + ( 2 x 2 − 1 ) e x 2 2 x 2 =-\frac{1}{2}x^\frac{-3}{2}-(x+1)e^x+10^xln10+\frac{(2x^2-1)e^{x^2}}{2x^2} = − 2 1 x 2 − 3 − ( x + 1 ) e x + 1 0 x l n 10 + 2 x 2 ( 2 x 2 − 1 ) e x 2 ;
(b) F ( x ) = e x s i n x + s e c x + l n x 2 ; F(x)=e^xsinx+secx+lnx^2; F ( x ) = e x s in x + sec x + l n x 2 ;
d F d x = e x s i n x + e x c o s x + t a n x c o s x + 2 x = \frac{dF}{dx}=e^xsinx+e^xcosx+\frac{tanx}{cosx}+\frac{2}{x}= d x d F = e x s in x + e x cos x + cos x t an x + x 2 =
= ( s i n x + c o s x ) e x + t a n x c o s x + 2 x ; =(sinx+cosx)e^x+\frac{tanx}{cosx}+\frac{2}{x}; = ( s in x + cos x ) e x + cos x t an x + x 2 ;
2.
We use the implicit function differentiation rule:
5 y 4 d y d x + 4 y 3 + 12 x y 2 d y d x + 3 x 2 = 0 ; 5y^4\frac{dy}{dx}+4y^3+12xy^2\frac{dy}{dx}+3x^2=0; 5 y 4 d x d y + 4 y 3 + 12 x y 2 d x d y + 3 x 2 = 0 ;
d y d x ( 5 y 4 + 12 x y 2 ) = − ( 4 y 3 + 3 x 2 ) ; \frac{dy}{dx}(5y^4+12xy^2)=-(4y^3+3x^2); d x d y ( 5 y 4 + 12 x y 2 ) = − ( 4 y 3 + 3 x 2 ) ;
d y d x = − 4 y 3 + 3 x 2 5 y 4 + 12 x y 2 . \frac{dy}{dx}=-\frac{4y^3+3x^2}{5y^4+12xy^2} . d x d y = − 5 y 4 + 12 x y 2 4 y 3 + 3 x 2 .
At the point (x,y)=(0,-1) we have:
y d x = − 4 ⋅ ( − 1 ) 3 + 0 5 ⋅ ( − 1 ) 4 + 0 = 4 5 . \frac{y}{dx}=-\frac{4\cdot (-1)^3+0}{5\cdot (-1)^4+0}=\frac{4}{5} . d x y = − 5 ⋅ ( − 1 ) 4 + 0 4 ⋅ ( − 1 ) 3 + 0 = 5 4 .
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