How do you differentiate #y=5x^6-sec^-1x#?

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How do you differentiate #y=5x^6-sec^-1x#?

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Answer 1 · 2018-02-13T08:02:34

#dy/dx=30x^5-1/(xsqrt(x^2-1)#

Explanation:

#y=5x^6-sec^-1x#
#dy/dx=d/dx(5x^6-sec^-1x)#
Using sum rule
#=d/dx(5x^6)-d/dx(sec^-1x)#
#d/dx(5x^6)=5.d/dx(x^6)#
#=(5)(6)x^(6-1)=30x^5#
#d/dx(sec^-1x)=1/(xsqrt(x^2-1)#
Thus,
#dy/dx=30x^5-1/(xsqrt(x^2-1)#

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How do you differentiate #y=5x^6-sec^-1x#?

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