Answer:
dy/dx = (cos x)^(log x)[(1/x)log(cosx) + (log x)(tanx)] + (log x)^x[log(logx) + (logx)^(-1)]
Step-by-step explanation:
Take, y1 = (cos x)^(log x)
Taking log both sides we get:
log(y1) = log[(cos x)^(log x)]
or,log(y1) = log[(cos x)^(log x)]
or,log(y1) = (log x)log(cos x)
Differentiating both sides w.r.t x we get:
(1/y1) dy1/dx = (1/x)log(cosx) + (log x)(-sinx/cosx)
(1/y1) dy1/dx = (1/x)log(cosx) + (log x)(-tanx)
dy1/dx = (cos x)^(log x)[(1/x)log(cosx) (log x)(tanx)]
Now take, y2 = (log x)^x
log(y2) = xlog(logx)
(1/y2)dy2/dx = log(logx) + (logx)^(-1)
dy2/dx = (log x)^x[log(logx) + (logx)^(-1)]
Therefore, dy/dx = dy1/dx + dy2/dx
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Verified answer
Answer:
dy/dx = (cos x)^(log x)[(1/x)log(cosx) + (log x)(tanx)] + (log x)^x[log(logx) + (logx)^(-1)]
Step-by-step explanation:
Take, y1 = (cos x)^(log x)
Taking log both sides we get:
log(y1) = log[(cos x)^(log x)]
or,log(y1) = log[(cos x)^(log x)]
or,log(y1) = (log x)log(cos x)
Differentiating both sides w.r.t x we get:
(1/y1) dy1/dx = (1/x)log(cosx) + (log x)(-sinx/cosx)
(1/y1) dy1/dx = (1/x)log(cosx) + (log x)(-tanx)
dy1/dx = (cos x)^(log x)[(1/x)log(cosx) (log x)(tanx)]
Now take, y2 = (log x)^x
Taking log both sides we get:
log(y2) = xlog(logx)
Differentiating both sides w.r.t x we get:
(1/y2)dy2/dx = log(logx) + (logx)^(-1)
dy2/dx = (log x)^x[log(logx) + (logx)^(-1)]
Therefore, dy/dx = dy1/dx + dy2/dx