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Find dy/dx: y=e^(5-3lnx)
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I want to know how to differentiate 5-3lnx
For any dy/dx ln(x) = 1/x * f'(x)
differentiation of 5 will be 0 and -3lnx will be -3/x
Ah, thought of a function, haha
\[\large \frac{d}{dx}\ln(x)=\frac{1}{x}\]
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yeah right:)
\[\large \frac{d}{dx}(5-3\ln x)=\frac{d}{dx}5-\frac{d}{dx}3\ln x=0-\frac{3}{x}=-\frac{3}{x}\]
\[\large \frac{d}{dx}(e)^{f(x)} = (e)^{f(x)} \times \frac{d}{dx}(f(x))\]
[e ^{5-3lnx}*(-3/x)\]
Yes you are right..
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Correct
Get it thank you so much, i was just confused about what rule to use.
Give the medal to zepp.. @jainilk
Apply the chain rule for this one
I thought jainik asked the question.. Ha ha ha..
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