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What is the derivative of y= x⁵ + cotx |
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Answer» y=ln(cosecx−cotx) Chain RULE, dx dy
= du dy
⋅ dx du
Here u=cosecx−cotx ⇒ dx dy
= cosecx−cotx 1
⋅ dx d(cosecx−cotx)
⇒ dx dy
= cosecx−cotx −cosecx.cotx+cosec
⇒ dx dy
= cosecx−cotx −cosecx(cotx−cosecx)
⇒ dx dy
=cosecx |
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