wyznacz pochodną po dz i dw:
\(\displaystyle{ \frac{4Pl^{3}}{3\pi(d_{z}^{4}-d_{w}^{4})h}}\)
dzieki
pochodna
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soku11
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- Rejestracja: 16 sty 2007, o 19:42
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pochodna
\(\displaystyle{ \frac{\mbox{d}}{\mbox{d}_z} ft(\frac{4Pl^3}{3\pi(d_{z}^{4}-d_{w}^{4})h}\right)=
\frac{4Pl^3}{3\pi h} ft(\frac{1}{(d_z^4-d_w^4}\right)'_{d_z}=
\frac{4Pl^3}{3\pi h} ft[(d_z^4-d_w^4)^{-1}\right]'_{d_z}=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (d_z^4-d_w^4)'_{d_z}=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (4d_z^3)=
\frac{-16Pl^3d_z^3}{3\pi h(d_z^4-d_w^4)^2}\\
\frac{\mbox{d}}{\mbox{d}_w} ft(\frac{4Pl^3}{3\pi(d_{z}^{4}-d_{w}^{4})h}\right)=(\ldots)=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (-4d_w^3)=
\frac{16Pl^3d_w^3}{3\pi h(d_z^4-d_w^4)^2}}\)
POZDRO
\frac{4Pl^3}{3\pi h} ft(\frac{1}{(d_z^4-d_w^4}\right)'_{d_z}=
\frac{4Pl^3}{3\pi h} ft[(d_z^4-d_w^4)^{-1}\right]'_{d_z}=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (d_z^4-d_w^4)'_{d_z}=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (4d_z^3)=
\frac{-16Pl^3d_z^3}{3\pi h(d_z^4-d_w^4)^2}\\
\frac{\mbox{d}}{\mbox{d}_w} ft(\frac{4Pl^3}{3\pi(d_{z}^{4}-d_{w}^{4})h}\right)=(\ldots)=
\frac{4Pl^3}{3\pi h}\cdot ft[-1(d_z^4-d_w^4)^{-2}\right]\cdot (-4d_w^3)=
\frac{16Pl^3d_w^3}{3\pi h(d_z^4-d_w^4)^2}}\)
POZDRO