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What is dydx of sincostanx?

Updated: 12/13/2022

Wiki User

∙ 16y ago

Best Answer

If y = sin(cos(tan(x)))

Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x)

Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x)

= -cos(cos(tan(x))).sin(tan(x)).sec2(x)

Unfortunately I don't think this can be simplified much more.

( sec = 1/cos )

Wiki User

∙ 16y ago

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What is Dydx of x(x-10)?

Assuming y = x(x - 10) Either use the product rule: u = x v = (x - 10) → y = x(x - 10) = uv d/dx (uv) = v du/dx + u dv/dx → dy/dx = d/dx x(x - 10) = (x - 10)(d/dx x) + (x)(d/dx (x - 10)) = (x - 10)(1) + (x)(1) = x - 10 + x = 2x - 10 or expand the brackets and differentiate dy/dx = d/dx x(x - 10) = d/dx x² - 10x = 2x - 10

Related questions

What is Dydx of x(x-10)?

Assuming y = x(x - 10) Either use the product rule: u = x v = (x - 10) → y = x(x - 10) = uv d/dx (uv) = v du/dx + u dv/dx → dy/dx = d/dx x(x - 10) = (x - 10)(d/dx x) + (x)(d/dx (x - 10)) = (x - 10)(1) + (x)(1) = x - 10 + x = 2x - 10 or expand the brackets and differentiate dy/dx = d/dx x(x - 10) = d/dx x² - 10x = 2x - 10

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