how to solve log(sin+tanx) graphically????\
Answers
Assume y≠log(sinx+tanx).
Since y is a sum, you can differentiate one piece at a time and then combine the answers.
ddy(y)=dydx.
ddxlogesinx
=1sinxddxsinx
=(1sinx)cosx
=cosxsinx
=cotx
from the formula ddxlogeu=1ududx.
ddxtanx
=ddxsinxcosx=ddx(sinx)(cosx)−1
=sinxddx(cosx)−1
+(cosx)−1ddxsinx
=(sinx)(−1)(cosx)−2ddxcosx
+(cosx)−1(cosx)
=(sinx)(−1)(cosx)−2(−sinx)+(cosx)−1(cosx)
=sin2xcos2x+1
=sin2xcos2x+cos2xcos2x
=sin2x+cos2xcos2x
=1cos2x=12cos2x
=sec2x.
Combining the answers, get
dydx=cotx+sec2x.
Here we need to understand that sin(x) graph is being put into (superimposed) the graph of log(x).
Green one is sin(x) and blue one is log(x).
We know that graph of sin(x) oscillates between -1 and 1. Also, graph of log(x) gives good negative values as x approaches 0 from right. Thus when sin(x) has a negative value the log(x) becomes undefined i.e, from π to 2π sin(x) has negative value. Then graph
What is the graph of log (sinx)?
How do you plot sin (log x)?
How do I find the period of sin (log x)?
How do I draw a sin x*logx graph?
What is the graph of sgn (sinx) and sgn (log x)?
y=log(sinx)
e^y=sinx
2.7182^y=sinx
But 2.7182^y > 0 for y>1, ………….(1)
2.7182^y = 1 for y=0 and …………..(2)
0<2.7182^y < 1 for y<0 …………(3)
Therefore, condition (1) will be rejected as sinx can't be greater than 1 and conditions will be accepted.
Hence, y≤0
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