x 2 − y 2 + 3 x = 5 y ⇒ 2 x − 2 y y ′ + 3 = 5 y ′ ⇒ 2 x + 3 = 2 y y ′ + 5 y ′ ⇒ 2 x + 3 = y ′ ( 2 y + 5 ) ⇒ y ′ = 2 x + 3 2 y + 5 {\displaystyle x^{2}-y^{2}+3x=5y{\Rightarrow }2x-2yy'+3=5y'{\Rightarrow }2x+3=2yy'+5y'{\Rightarrow }2x+3=y'(2y+5){\Rightarrow }y'={\frac {2x+3}{2y+5}}}
α) y = f ( x ) ⇒ y ′ = f ′ ( x ) 2 f ( x ) {\displaystyle y={\sqrt {f(x)}}{\Rightarrow }y'={\frac {f'(x)}{2{\sqrt {f(x)}}}}} β) y = κ x ν ⇒ y ′ = − κ ∗ ν x ν + 1 {\displaystyle y={\frac {\kappa }{x^{\nu }}}{\Rightarrow }y'=-{\frac {{\kappa }*{\nu }}{x^{{\nu }+1}}}} γ) y = f ( x ) ν α { y ′ = ν α ∗ f ′ ( x ) f ( x ) α − ν α , γ ι α ν < α y ′ = ν α ∗ f ( x ) ν − α α ∗ f ′ ( x ) , γ ι α ν > α {\displaystyle {y=f(x)^{{\nu } \over {\alpha }}{\begin{cases}y'={{\nu } \over {\alpha }}*{\frac {f'(x)}{\sqrt[{\alpha }]{f(x)^{{\alpha }-{\nu }}}}},&\gamma \iota \alpha \;{\nu }<{\alpha }\;\\y'={{\nu } \over {\alpha }}*{\sqrt[{\alpha }]{f(x)^{{\nu }-{\alpha }}}}*f'(x),&\gamma \iota \alpha \;{\nu }>{\alpha }\;\end{cases}}}} δ) y = α f ( x ) ν ⇒ y ′ = f ′ ( x ) ∗ α f ( x ) l n α ν {\displaystyle y={\frac {{\alpha }^{f(x)}}{\nu }}{\Rightarrow }y'={\frac {f'(x)*{\alpha }^{f(x)}ln{\alpha }}{\nu }}}
επιστροφή στο τμήμα μαθηματικών.
![{\displaystyle {y=f(x)^{{\nu } \over {\alpha }}{\begin{cases}y'={{\nu } \over {\alpha }}*{\frac {f'(x)}{\sqrt[{\alpha }]{f(x)^{{\alpha }-{\nu }}}}},&\gamma \iota \alpha \;{\nu }<{\alpha }\;\\y'={{\nu } \over {\alpha }}*{\sqrt[{\alpha }]{f(x)^{{\nu }-{\alpha }}}}*f'(x),&\gamma \iota \alpha \;{\nu }>{\alpha }\;\end{cases}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85e579c26c6e7a66d9e4cff4e6ee4eea23591694)
