Question

Differentiate sinX w. r. t cosX

Answer 1

Ans. is (-cotx)
Let's differentiate sinx with respect to x
d(sinx)/dx= cosx............(I).
again' differentiate cosx with respect to x
d(cosx)/dx = -sinx............(ii)

Now we know that,
d(sinx)/d(cosx)= {d(sinx)/dx}/{d(cosx)/dx }

d(sinx)/d(cosx)= cosx/(-sinx) .......form euqation (I) and (ii).
d(sinx)/d(cosx)= -cotx
Hope it helped you!!!

Answer 2

◢LET'S

=> u = sinx

◢AND

=> v = cosx

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◢THEN,

$= > u = sinx \\ \\ = > \frac{du}{dx} = cosx \: \: \: \: \: \: \: ..[Eq _{1}]$

◢AND

$= > v = cosx \\ \\ = > \frac{dv}{dx} = - sinx \: \: \: \: \: ....[Eq _{2}]$
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◢SO,

● [Eq(1) ÷ Eq(2)]

◢THEN,

$= > \frac{du}{dx} \div \frac{dv}{dx} = \frac{cosx}{ - sinx} \\ \\ = > \frac{du}{ dx} \times \frac{dx}{dv} = - \frac{cosx}{sinx} \\ \\ = > \frac{du}{dv} = - cotx \: \: \: \: ...[ANSWER]$
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