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How do we differentiate a^x? - Example 31 - Chapter 5 Class 12

Example 31 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Example 31 Differentiate ๐‘Ž^๐‘ฅ ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ, where a is a positive constant.Let y = ๐‘Ž^๐‘ฅ Taking log on both sides logโก๐‘ฆ = logโก๐‘Ž^๐‘ฅ ๐’๐’๐’ˆโก๐’š = ๐’™ ๐’๐’๐’ˆโก ๐’‚ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฅ = ๐‘‘/๐‘‘๐‘ฅ(๐‘ฅ logโก๐‘Ž) (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฅ = logโก๐‘Ž (๐‘‘๐‘ฅ/๐‘‘๐‘ฅ) (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฅ = ๐’๐’๐’ˆโก๐’‚ (๐‘™๐‘œ๐‘”โกใ€–๐‘Ž^๐‘=๐‘ ๐‘™๐‘œ๐‘”โก๐‘Ž ใ€—) (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ = logโก๐‘Ž (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = logโก๐‘Ž 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = logโก๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ logโก๐‘Ž Putting back ๐‘ฆ = ๐‘Ž^๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐’‚^๐’™ ๐’๐’๐’ˆโก๐’‚

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