Let's differentiate: $\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt[8]{x}-8e^x)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt[8]{x}-8e^x)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt[8]{x}\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-8e^x)$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt[8]{x}\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-8e^x)$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt[8]{x}\right)$ = $\frac{1}{8}·x^{-\frac{7}{8}}$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}(-8e^x)$ = $-8·\frac{\mathrm{d}}{\mathrm{d}x}\left(e^x\right)$■The derivative of exp(x) is exp(x): $\frac{\mathrm{d}}{\mathrm{d}x}\left(e^x\right)$ = $e^x$

$\frac{1}{8}·x^{-\frac{7}{8}}-8e^x$

ⓘ Multiplying a fraction by a term multiplies its numerator: $\frac{1}{8}·x^{-\frac{7}{8}}$ = $\frac{1·x^{-\frac{7}{8}}}{8}$ = $\frac{x^{-\frac{7}{8}}}{8}$■The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base: $x^{-\frac{7}{8}}$ = $\frac{1}{\sqrt[8]{x^7}}$

$\frac{\frac{1}{\sqrt[8]{x^7}}}{8}-8e^x$

ⓘ Dividing a fraction by $8$ is multiplying its denominator by $8$: $\frac{\frac{1}{\sqrt[8]{x^7}}}{8}$ = $\frac{1}{8\sqrt[8]{x^7}}$

▶Derivative:   $\frac{1}{8\sqrt[8]{x^7}}-8e^x$