calculas differentiation 1) d/dx = ln(10x^3+6) 2) y=e^(2t^3+6) find dy/dt

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\frac{d y}{d x}

3 x^{2} + t a n x + s e c x = y + 6

y = 3 \frac{d y}{d x} + 6 {(\frac{d y}{d x})}^{2} y^{2}

\frac{d y}{d x}

\frac{d x}{d t} = a p cos p t, \frac{d y}{d t} = - b p sin p t

\frac{d^{2} y}{{d x}^{2}} = {[y + {(\frac{d y}{d x})}^{6}]}^{1 / 4}

{[5 + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = 6 \frac{d^{2} y}{d x^{2}}

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