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Differentiate
y=tan (x^2-5x+6)​

User Jan Joswig
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1 Answer

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Use the chain rule:

y = tan(x ² - 5x + 6)

y' = sec²(x ² - 5x + 6) × (x ² - 5x + 6)'

y' = (2x - 5) sec²(x ² - 5x + 6)

Perhaps more explicitly: let u(x) = x ² - 5x + 6, so that

y(x) = tan(x ² - 5x + 6) → y(u(x)) = tan(u(x) )

By the chain rule,

y'(x) = y'(u(x)) × u'(x)

and we have

y(u) = tan(u) → y'(u) = sec²(u)

u(x) = x ² - 5x + 6 → u'(x) = 2x - 5

Then

y'(x) = (2x - 5) sec²(u)

or

y'(x) = (2x - 5) sec²(x ² - 5x + 6)

as we found earlier.

User Ashish Shukla
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