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Calculate the derivative of the function y = log(2x) base 10

To find the derivative of the function y=log10(2x)y = \log_{10}(2x), we can apply the chain rule. We start by letting u=2xu = 2x, which allows us to rewrite the function as y=log10(u)y = \log_{10}(u).

By applying the chain rule, we get:

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Next, we need to determine dydu\frac{dy}{du}. The derivative of a logarithmic function is given by the formula:

dduloga(u)=1ulna\frac{d}{du} \log_a(u) = \frac{1}{u \ln a}

In this case, since a=10a = 10, we have:

dydu=1uln10\frac{dy}{du} = \frac{1}{u \ln 10}

Now, substituting back for uu, we find:

dydu=1(2x)ln10\frac{dy}{du} = \frac{1}{(2x) \ln 10}

Next, we compute dudx\frac{du}{dx}:

dudx=ddx(2x)=2\frac{du}{dx} = \frac{d}{dx}(2x) = 2

Now, substituting dydu\frac{dy}{du} and dudx\frac{du}{dx} back into our chain rule equation, we have:

dydx=1(2x)ln102\frac{dy}{dx} = \frac{1}{(2x) \ln 10} \cdot 2

Simplifying this expression, we find:

dydx=2(2x)ln10=1xln10\frac{dy}{dx} = \frac{2}{(2x) \ln 10} = \frac{1}{x \ln 10}

Thus, the derivative of y=log10(2x)y = \log_{10}(2x) is:

dydx=1xln10\frac{dy}{dx} = \frac{1}{x \ln 10}
Answered by: Prof. Sophia Clark
A-Level Maths Tutor
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