Question

write the derivative of the following​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

We know that,

$\frac{d(x^n)}{dx} = n\cdot x^{n-1} , \frac{d{(cos{x})}}{dx} = -sin{x} ,\\\frac {d{(u\cdot v)}}{dx} = u\cdot \frac{dv}{dx} +v\cdot \frac{du}{dx}, \frac{d(tan{x})}{dx} = sec^{2}{x} \text{ and } \frac{d(constant)}{dx} = 0$

Now,

$y = \frac{1}{x^2} + \frac{2}{x^3} + \frac{3}{x^4} +\frac{4}{x^5} - 4cos{x} - 3x\cdot tan{x} + 7$

$or,y = x^{-2} +2\cdot x^{-3} + 3\cdot x^{-4} +4\cdot x^{-5} - 4cos{x} - 3x\cdot tan{x}+7 \\or,\frac{dy}{dx} = 1\cdot (-2)\cdot x^{-2-1} + 2\cdot (-3)\cdot x^{-3-1} + 3\cdot (-4)\cdot x^{-4-1} +4\cdot (-5)\cdot x^{-5-1} - 4\cdot (-sin{x}) - (3\cdot tan{x} +3x.sec^{2} {x} ) + 0\\or, \frac{dy}{dx} = (-2)\cdot x^{-3} +(-6)\cdot x^{-4} + (-12)\cdot x^{-5} +(-20)\cdot x^{-6} + 4sin{x} - 3\cdot tan{x} - 3x.sec^{2} {x}$

$or, \frac{dy}{dx} = -2\cdot x^{-3} -6\cdot x^{-4} -12\cdot x^{-5} -20\cdot x^{-6} + 4sin{x} - 3\cdot tan{x} - 3x.sec^{2} {x}$